Chapter 04 Review Problems

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The Mole and Molar Mass

Which sample contains the most particles?

1. 1 mol H₂O(g)
2. 1 mol UF₆(g)
3. 1 mol CH₃COOH(l)
4. 1 mol He(g)
5. all contain the same number of particles

Solution

Answer: E

Concept: number of entities; moles

Since there is 1 mole of each sample, each sample contains an Avogadro number (6.022… × 10²³) of particles.

Which sample contains the most atoms?

1. 1 mol CO₂(g)
2. 1 mol UF₆(g)
3. 1 mol CH₃COCH₃(l)
4. 1 mol He(g)
5. all contain the same number of particles

Solution

Answer: C

Concept: moles; atomic composition

To find the total number of atoms in each sample, we must multiply the number of moles of the substance by the number of atoms in each molecule or formula unit. Since all samples are 1 mol, the sample with the most atoms per molecule will contain the most total atoms. Let n(atoms) represent the total moles of atoms.

1. 1 mol H₂O: \[ \begin{align*} n(\mathrm{atoms}) &= n(\mathrm{H_2O}) ~ r(\mathrm{atoms,H_2O}) \\[1.5ex] &= \left( 1~\mathrm{mol~H_2O} \right) \left( \frac{3~\mathrm{mol~atoms}}{1~\mathrm{mol~H_2O}} \right) \\[1.5ex] &= 3~\mathrm{mol~atoms} \end{align*} \]

2. 1 mol UF₆: \[ \begin{align*} n(\mathrm{atoms}) &= n(\mathrm{UF_6}) ~ r(\mathrm{atoms,UF_6}) \\[1.5ex] &= \left( 1~\mathrm{mol~UF_6} \right) \left( \frac{7~\mathrm{mol~atoms}}{1~\mathrm{mol~UF_6}} \right) \\[1.5ex] &= 7~\mathrm{mol~atoms} \end{align*} \]

3. 1 mol CH₃COCH₃: \[ \begin{align*} n(\mathrm{atoms}) &= n(\mathrm{CH_3COCH_3}) ~ r(\mathrm{atoms,CH_3COCH_3}) \\[1.5ex] &= \left( 1~\mathrm{mol~CH_3COCH_3} \right) \left( \frac{10~\mathrm{mol~atoms}}{1~\mathrm{mol~CH_3COCH_3}} \right) \\[1.5ex] &= 10~\mathrm{mol~atoms} \end{align*} \]

4. 1 mol He: \[ \begin{align*} n(\mathrm{atoms}) &= n(\mathrm{He}) ~ r(\mathrm{atoms,He}) \\[1.5ex] &= \left ( 1~\mathrm{mol~He} \right ) \left( \frac{1~\mathrm{mol~atom}}{1~\mathrm{mol~He}} \right) \\[1.5ex] &= 1~\mathrm{mol~atoms} \end{align*} \]

Comparing the results, the 1 mole sample of acetone (CH₃COCH₃) contains the most moles of atoms, and therefore the most total atoms.

How many atoms (in normalized scientific notation) are present in a 0.268 mol sample of CH₃OH?

Solution

Answer: 9.68 × 10²³ atoms

Concept: number of particles; moles

Every molecule of CH₃OH contains 6 atoms; therefore, every 1 mol of CH₃OH contains 6 mol of atoms.

\[ \begin{align*} N(\mathrm{atoms}) &= n(\mathrm{atoms}) ~ N_{\mathrm{A}} \\[1.5ex] &= n(\mathrm{CH_3OH}) ~ r(\mathrm{atoms,CH_3OH}) ~ N_{\mathrm{A}} \\[1.5ex] &= 0.26\bar{8}~\mathrm{mol~CH_3OH} \left ( \dfrac{6~\mathrm{mol~atoms}}{\mathrm{mol~CH_3OH}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}~\mathrm{atoms}}{\mathrm{mol~atoms}} \right ) \\[1.5ex] &= 9.6\bar{8}33\times 10^{23} \\[1.5ex] &= 9.68\times 10^{23} \end{align*} \]

Which one of the samples has the largest mass?

1. 1 mol CO₂(g)
2. 1 mol UF₆(g)
3. 1 mol CH₃COOH(l)
4. 1 mol He(g)
5. all contain the same number of particles

Solution

Answer: B

Concept: moles; molar mass

The mass of one mole of any substance is its molar mass (M), which is calculated by summing the atomic masses of all atoms in its chemical formula. To find the sample with the largest mass, we need to find the substance with the largest molar mass.

Let’s calculate the approximate molar mass for each substance:

1. CO₂: 1(12.01) + 2(16.00) = 44.01 g mol⁻¹
2. UF₆: 1(238.03) + 6(19.00) = 352.03 g mol⁻¹
3. CH₃COOH: 2(12.01) + 4(1.01) + 2(16.00) = 60.06 g mol⁻¹
4. He: 4.00 g mol⁻¹
5. This answer option does not answer the question.

Comparing the approximate molar masses, UF₆ is by far the most massive. Therefore, one mole of UF₆ has the largest mass.

How many aluminum atoms are there in a 3.50 g sample of Al₂O₃?

Solution

Answer: 4.13 × 10²² atoms

Concept: number of particles; moles

The molar mass of Al₂O₃ is 101.9 g mol⁻¹.

\[ \begin{align*} N(\mathrm{Al}) &= n(\mathrm{Al}) ~ N_{\mathrm{A}} \\[1.5ex] &= n(\mathrm{Al_2O_3}) ~ r(\mathrm{Al,Al_2O_3}) ~ N_{\mathrm{A}} \\[1.5ex] &= m(\mathrm{Al_2O_3}) ~ M(\mathrm{Al_2O_3})^{-1} ~ r(\mathrm{Al,Al_2O_3}) ~ N_{\mathrm{A}} \\[1.5ex] &= 3.5\bar{0}~\mathrm{g~Al_2O_3} \left ( \dfrac{\mathrm{mol~Al_2O_3}}{101.9\bar{6}~\mathrm{g}} \right ) \left ( \dfrac{2~\mathrm{mol~Al}}{\mathrm{mol~Al_2O_3}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}~\mathrm{Al}}{\mathrm{mol~Al}} \right ) \\[1.5ex] &= 4.1\bar{3}43\times 10^{22} \\[1.5ex] &= 4.13\times 10^{22} \end{align*} \]

What is the mass (in g) of 3.6 × 10²² chlorine atoms?

Solution

Answer: 2.1 g

Concept: the mole; the Avogadro constant

To find the mass, we must first convert the number of atoms (N) to the amount in moles (n) using Avogadro’s constant (N_A). Then, we convert from moles to mass using the molar mass (M).

\[ \begin{align*} m(\mathrm{Cl}) &= n(\mathrm{Cl}) ~ M(\mathrm{Cl})\\[1.5ex] &= N(\mathrm{Cl}) ~ N_{\mathrm{A}}{^{-1}} ~ M(\mathrm{Cl}) \\[1.5ex] &= 3.\bar{6}\times 10^{22}~\mathrm{Cl} ~ \left ( \dfrac{\mathrm{mol}}{6.02\bar{2}\times 10^{23}~\mathrm{Cl}} \right ) \left ( \dfrac{35.4\bar{5}~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 2.\bar{1}16~\mathrm{g} \\[1.5ex] &= 2.1~\mathrm{g} \end{align*} \]

How many iron atoms (in normalized scientific notation) are contained in 354 g of iron?

Solution

Answer: 3.82 × 10²⁴ atoms

Concept: moles; the Avogadro constant

\[ \begin{align*} N(\mathrm{Fe}) &= n(\mathrm{Fe}) ~ M(\mathrm{Fe}) \\[1.5ex] &= m(\mathrm{Fe}) ~ M(\mathrm{Fe})^{-1} ~ N_{\mathrm{A}} \\[1.5ex] &= 35\bar{4}~\mathrm{g} ~ \left ( \dfrac{\mathrm{mol}}{55.8\bar{5}~\mathrm{g}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}~\mathrm{Fe}}{\mathrm{mol}} \right ) \\[1.5ex] &= 3.8\bar{1}69 \times 10^{24}~\mathrm{Fe}\\[1.5ex] &= 3.82\times 10^{24}~\mathrm{Fe} \end{align*} \]

What is the molar mass (in g mol^–1) of Al₂(SO₄)₃ · 18 H₂O?

Solution

Answer: 666.50 g mol^–1

Concept: molar mass

\[ \begin{align*} M(\mathrm{Al_2(SO_4)_3 \boldsymbol{\cdot} 18~H_2O}) &= M_{\mathrm{r}} ~ [\mathrm{Al_2(SO_4)_3 \boldsymbol{\cdot} 18~H_2O}] ~ M_{\mathrm{u}} \\[1.5ex] &= [ 2~A_{\mathrm{r}}{^{\circ}}(\mathrm{Al}) + 3~A_{\mathrm{r}}{^{\circ}}(\mathrm{S}) + 30~A_{\mathrm{r}}{^{\circ}}(\mathrm{O}) + 36~A_{\mathrm{r}}{^{\circ}}(\mathrm{H}) ] ~ M_{\mathrm{u}} \\[1.5ex] &= [ 2 \left ( 26.9\bar{8} \right ) + 3 \left ( 32.0\bar{6} \right ) + 30 \left ( 16.0\bar{0} \right ) + 36 \left ( 1.0\bar{1} \right ) ] \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= [ 53.9\bar{6}0 + 96.1\bar{8}0 + 480.0\bar{0}0 + 36.3\bar{6}0 ] \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 666.5\bar{0}0~\mathrm{g~mol^{-1}} \\[1.5ex] &= 666.50~\mathrm{g~mol^{-1}} \end{align*} \]

What is the molar mass (in g mmol^–1) of (NH₄)₂HPO₄?

Solution

Answer: 0.13208 g mmol^–1

Concept: molar mass

\[ \begin{align*} M[\mathrm{(NH_4)_2HPO_4}] &= M_{\mathrm{r}}[\mathrm{(NH_4)_2HPO_4}] ~ M_{\mathrm{u}} \\[1.5ex] &= [ 2~A_{\mathrm{r}}{^{\circ}}(\mathrm{N}) + 9~A_{\mathrm{r}}{^{\circ}}(\mathrm{H}) + A_{\mathrm{r}}{^{\circ}}(\mathrm{P}) + 4~A_{\mathrm{r}}{^{\circ}}(\mathrm{O}) ] ~ M_{\mathrm{u}} \\[1.5ex] &= [ 2 \left ( 14.0\bar{1} \right ) + 9 \left ( 1.0\bar{1} \right ) + \left ( 30.9\bar{7} \right ) + 4 \left ( 16.0\bar{0} \right ) ] \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{mol}}{10^3~\mathrm{mmol}} \right ) \\[1.5ex] &= [ 28.0\bar{2}0 + 9.0\bar{9}0 + 30.9\bar{7}0 + 64.0\bar{0}0 ] \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{mol}}{10^3~\mathrm{mmol}} \right )\\[1.5ex] &= 132.0\bar{8}0 \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{mol}}{10^3~\mathrm{mmol}} \right )\\[1.5ex] &= 0.1320\bar{8}0~\mathrm{g~mmol^{-1}} \\[1.5ex] &= 0.13208~\mathrm{g~mmol^{-1}} \end{align*} \]

What is the mass (in ng as normalized scientific notation) of 2.44 × 10²⁰ oxygen atoms?

Solution

Answer: 6.48 × 10⁶ ng

Concept: moles; the Avogadro constant

\[ \begin{align*} m(\mathrm{O}) &= n(\mathrm{O}) ~ M(\mathrm{O}) \\[1.5ex] &= N(\mathrm{O}) ~ N_{\mathrm{A}}^{-1} ~ M(\mathrm{O}) \\[1.5ex] &= 2.4\bar{4}\times 10^{20}~\mathrm{O} ~ \left ( \dfrac{\mathrm{mol}}{6.02\bar{2}\times 10^{23}~\mathrm{O}} \right ) \left ( \dfrac{16.0\bar{0}~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{10^9~\mathrm{ng}}{\mathrm{g}} \right ) \\[1.5ex] &= 6.4\bar{8}28\times 10^{6}~\mathrm{ng} \\[1.5ex] &= 6.48\times 10^{6}~\mathrm{ng} \end{align*} \]

A sample of a compound containing only carbon and oxygen decomposes and produces 24.5 g of carbon and 32.59 g of oxygen. What is the sample?

1. CO
2. CO₂
3. CO₃
4. C₃O₂
5. C₂O

Solution

Answer: A

Concept: Law of Definite Proportions; empirical formula; percent by mass; mole ratio

The most direct way to identify the compound is to determine its empirical formula from the experimental masses. The empirical formula represents the simplest whole-number ratio of moles of each element in the compound.

Step 1: Convert the Mass of Each Element to Moles \[ \begin{align*} n(\mathrm{C}) &= m(\mathrm{C}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \left( 24.\bar{5}~\mathrm{g~C} \right) \left( \frac{1~\mathrm{mol~C}}{12.0\bar{1}~\mathrm{g~C}} \right) \\[1.5ex] &= 2.0\bar{3}99~\mathrm{mol~C} \\[2ex] n(\mathrm{O}) &= m(\mathrm{O}) ~ M(\mathrm{O})^{-1} \\[1.5ex] &= \left( 32.5\bar{9}~\mathrm{g~O} \right) \left( \frac{1~\mathrm{mol~O}}{16.0\bar{0}~\mathrm{g~O}} \right) \\[1.5ex] &= 2.03\bar{1}25~\mathrm{mol~O} \end{align*} \]

Step 2: Find the Simplest Whole-Number Ratio

Divide the mole amount of each element by the smallest mole amount (2.03125 mol) to find the ratio. \[ \begin{align*} \text{C Ratio} &= \frac{2.0\bar{3}99~\mathrm{mol}}{2.03\bar{1}25~\mathrm{mol}} \\[1.5ex] &= 1.0\bar{0}42 \\[1.5ex] &\approx 1 \\[1.5ex] \text{O Ratio} &= \frac{2.03\bar{1}25~\mathrm{mol}}{2.03\bar{1}25~\mathrm{mol}} \\[1.5ex] &= 1.00\bar{0}0 \\[1.5ex] &= 1 \end{align*} \] The mole ratio of C to O is 1:1. The empirical formula of the compound is CO.

---

An Intuitive Approach (without full calculation)

You are correct that you can often narrow down the choices without performing the full calculation.

1. Estimate the Mass Ratio: The problem gives 24.5 g C and 32.59 g O. These masses are roughly in a 24:32 ratio, which simplifies to 3:4.
2. Estimate the Molar Mass Ratio: The molar mass of Carbon is ~12 g mol⁻¹, and Oxygen is ~16 g mol⁻¹. This is also a 12:16 or 3:4 ratio.
3. Compare Ratios: Since the ratio of the masses in the sample is approximately the same as the ratio of the molar masses of the individual atoms, it strongly implies that the atoms are present in a 1-to-1 mole ratio. The only formula that fits this is CO.

The other options would have very different mass ratios. For example, in CO₂, you would expect the mass of oxygen to be significantly more than double the mass of carbon.

---

Alternative Method: Comparing Mass Percentages

Another, less direct way to solve the problem is to find the mass percent of each element in the sample and then compare it to the theoretical mass percent of each possible compound.

Total mass of compuond:

\[ \begin{align*} m(\mathrm{sample}) &= 24.\bar{5}~\mathrm{g} + 32.5\bar{9}~\mathrm{g} = 57.\bar{0}9~\mathrm{g} \end{align*} \]

Mass percent of C and O in sample:

\[ \begin{align*} w(\mathrm{C})~\% &= w(\mathrm{C}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{m(\textrm{C})}{m(\mathrm{sample})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{24.\bar{5}~\mathrm{g}}{57.\bar{0}9~\mathrm{g}} \right ) \times 100~\% \\[1.5ex] &= 42.\bar{9}14~\% \\[3ex] w(\mathrm{O})~\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{m(\textrm{O})}{m(\mathrm{sample})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{32.5\bar{9}~\mathrm{g}}{57.\bar{0}9~\mathrm{g}} \right ) \times 100~\% \\[1.5ex] &= 57.\bar{0}85~\% \end{align*} \]

Mass percent of C and O in CO:

\[ \begin{align*} w(\mathrm{C})~\% &= w(\mathrm{C}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{A_{\textrm{r}}^{\circ}(\textrm{C})} {M_{\mathrm{r}}(\textrm{CO})} \right ) \times 100~\% \\[1.5ex] &= \left (\dfrac{A_{\textrm{r}}^{\circ}(\textrm{C})} { \left [ A_{\textrm{r}}^{\circ}(\textrm{C}) + A_{\textrm{r}}^{\circ}(\textrm{O}) \right ] } \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{12.0\bar{1}} {\left ( 12.0\bar{1} + 16.0\bar{0} \right )} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{12.0\bar{1}}{28.0\bar{1}} \right ) \times 100~\% \\[3ex] &= 42.8\bar{7}75~\% \\[3ex] w(\mathrm{O})~\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{A_{\textrm{r}}^{\circ}(\textrm{O})} {M_{\mathrm{r}}(\textrm{CO})} \right ) \times 100~\% \\[1.5ex] &= \left (\dfrac{A_{\textrm{r}}^{\circ}(\textrm{O})} { \left [ A_{\textrm{r}}^{\circ}(\textrm{C}) + A_{\textrm{r}}^{\circ}(\textrm{O}) \right ] } \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{16.0\bar{0}} {\left ( 12.0\bar{1} + 16.0\bar{0} \right )} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{16.0\bar{0}}{28.0\bar{1}} \right ) \times 100~\% \\[1.5ex] &= 57.1\bar{2}24~\%\\[3ex] \end{align*} \]

The mass percent of C and O in the sample closely match the mass percent of C and O in CO. The other options give significantly different mass percent compositions. While this method works, it can be more time-consuming than the direct empirical formula calculation if the first guess is not correct.

How many moles (in normalized scientific notation) of Cs are contained in 595 kg of Cs?

Solution

Answer: 4.48 × 10³ mol

Concept: moles

\[ \begin{align*} n(\mathrm{Cs}) &= m(\mathrm{Cs}) ~ M(\mathrm{Cs})^{-1} \\[1.5ex] &= m(\mathrm{Cs}) ~ \left [ A_{\mathrm{r}}{^{\circ}}(\mathrm{Cs}) ~ M_{\mathrm{u}} \right ]^{-1} \\[1.5ex] &= 59\bar{5}~\mathrm{kg} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left [ \left ( 132.9\bar{1} \right ) \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \right ]^{-1} \\[1.5ex] &= 44\bar{7}6.7~\mathrm{mol} \\[1.5ex] &= 4.48\times 10^{3}~\mathrm{mol} \end{align*} \]

What is the mass (in g as normalized scientific notation) of 2.2 × 10²⁴ mercury atoms?

Solution

Answer: 7.3 × 10² g

Concept: moles; the Avogadro constant

\[ \begin{align*} m(\mathrm{Hg}) &= n(\mathrm{Hg}) ~ M(\mathrm{Hg}) \\[1.5ex] &= N(\mathrm{Hg}) ~ N_{\mathrm{A}}{^{-1}} ~ M(\mathrm{Hg}) \\[1.5ex] &= 2.\bar{2}\times 10^{24}~\mathrm{Hg} ~ \left ( \dfrac{\mathrm{mol}}{6.02\bar{2}\times 10^{23}~\mathrm{Hg}} \right ) \left ( \dfrac{200.5\bar{9}~\mathrm{g}}{\mathrm{mol}} \right )\\[1.5ex] &= 7.\bar{3}28\times 10^{2}~\mathrm{g} \\[1.5ex] &= 7.3\times 10^{2}~\mathrm{g} \end{align*} \]

How much Fe (in mol and number of atoms, the latter as scientific notation) are in 225.0 g of Fe?

Solution

Answer: 4.029 mol; 2.426 × 10²⁴ atoms

Concept: moles; the Avogadro constant

\[ \begin{align*} n(\mathrm{Fe}) &= m(\mathrm{Fe}) ~ M(\mathrm{Fe})^{-1} \\[1.5ex] &= 225.\bar{0}~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{55.8\bar{5}~\mathrm{g}} \right ) \\[1.5ex] &= 4.02\bar{8}64~\mathrm{mol} \\[1.5ex] &= 4.029~\mathrm{mol}\\[2.0ex] N(\mathrm{Fe}) &= n(\mathrm{Fe}) ~ N_{\mathrm{A}} \\[1.5ex] &= 4.02\bar{8}64~\mathrm{mol} \left ( \dfrac{6.02\bar{2}\times 10^{23}~\mathrm{Fe}} {\mathrm{mol}} \right ) \\[1.5ex] &= 2.42\bar{6}04\times 10^{24} \\[1.5ex] &= 2.426\times 10^{24} \end{align*} \]

Freon-12 (CCl₂F₂), a gas that has an alarming ozone depletion potential, was commonly used as a refrigerant in household air conditioners. A typical 3-ton AC unit requires around 6–12 lbs. of freon-12. Determine the following:

1. the number of freon-12 molecules (in normalized scientific notation) in 5.44 kg (12 lbs.) of freon-12
2. the mass (in kg) of Cl in a 5.44 kg sample of freon-12?

Solution

Answer: 2.71 × 10²⁵ molecules; 3.19 kg

Concept: moles; the Avogadro constant

1. Find the number of CCl₂F₂ molecules in a 5.44 kg sample.

\[ \begin{align*} N(\mathrm{CCl_2F_2}) &= n(\mathrm{CCl_2F_2}) ~ N_{\mathrm{A}} \\[1.5ex] &= m(\mathrm{CCl_2F_2}) ~ M(\mathrm{CCl_2F_2})^{-1} ~ N_{\mathrm{A}} \\[1.5ex] &= 5.4\bar{4}~\mathrm{kg} ~ \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{mg}} \right ) \left ( \dfrac{\mathrm{mol}}{120.9\bar{1}~\mathrm{g}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= 2.7\bar{0}94 \times 10^{25}~\mathrm{CCl_2F_2} \\[1.5ex] &= 2.71\times 10^{25}~\mathrm{CCl_2F_2} \end{align*} \]

2. Find the mass of Cl atoms in a 5.44 kg sample of CCl₂F₂.

\[ \begin{align*} m(\mathrm{Cl}) &= n(\mathrm{Cl}) ~ M(\mathrm{Cl}) \\[1.5ex] &= n(\mathrm{CCl_2F_2}) ~ r(\mathrm{Cl,CCl_2F_2}) ~ M(\mathrm{Cl}) \\[1.5ex] &= N(\mathrm{CCl_2F_2}) ~ N_{\mathrm{A}}{^{-1}} ~ r(\mathrm{Cl,CCl_2F_2}) ~ M(\mathrm{Cl}) \\[1.5ex] &= 2.7\bar{0}94 \times 10^{25}~\mathrm{CCl_2F_2} ~ \left ( \dfrac{\mathrm{mol}}{6.02\bar{2}\times 10^{23}~\mathrm{Cl}} \right ) \left ( \dfrac{2~\mathrm{mol~Cl}}{1~\mathrm{mol~CCl_2F_2}} \right ) \left ( \dfrac{35.4\bar{5}~\mathrm{g}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \\[2.0ex] &= 3.1\bar{8}99~\mathrm{kg} \\[1.5ex] &= 3.19~\mathrm{kg} \end{align*} \]

Prevacid (C₁₆H₁₄F₃N₃O₂S) is used to treat gastroesophageal reflux disease (GERD). Determine each of the following:

1. the molar mass (in g mol^–1) of Prevacid
2. the mass (in g) of fluorine in a 0.85 mol sample of Prevacid
3. the number of C atoms (in normalized scientific notation) in a 0.85 mol sample of Prevacid
4. the mass (in g) of 5.25 × 10²¹ molecules of Prevacid

Solution

Answer:

1. 369.39 g mol^–1
2. 49 g
3. 8.2 × 10²⁴
4. 3.22 g

Concept: moles, molar mass, and the Avogadro constant

A. Molar mass of Prevacid

\[ \begin{align*} M(\mathrm{Prevacid}) &= M_{\mathrm{r}}(\mathrm{Prevacid}) ~ M_{\mathrm{u}} \\[1.5ex] &= \Biggl\{ \left [ 16~A_{\mathrm{r}}{^{\circ}}(\mathrm{C}) \right ] + \left [ 14~A_{\mathrm{r}}{^{\circ}}(\mathrm{H}) \right ] + \left [ 3~A_{\mathrm{r}}{^{\circ}}(\mathrm{F}) \right ] + \left [ 3~A_{\mathrm{r}}{^{\circ}}(\mathrm{N}) \right ] + \left [ 2~A_{\mathrm{r}}{^{\circ}}(\mathrm{O}) \right ] + A_{\mathrm{r}}{^{\circ}}(\mathrm{S}) \Biggl\} ~ M_{\mathrm{u}} \\[1.5ex] &= \Biggl\{ \left [ 16 \times 12.0\bar{1} \right ] + \left [ 14 \times 1.0\bar{1} \right ] + \left [ 3 \times 19.0\bar{0} \right ] + \left [ 3 \times 14.0\bar{1} \right ] + \left [ 2 \times 16.0\bar{0} \right ] + 32.0\bar{6} \Biggl\} ~ \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 369.3\bar{9}0~\mathrm{g~mol^{-1}} \\[1.5ex] &= 369.39~\mathrm{g~mol^{-1}} \end{align*} \]

B. Mass (in g) of F in 0.75 mol Prevacid

\[ \begin{align*} m(\mathrm{F}) &= n(\mathrm{F}) ~ M(\mathrm{F}) \\[1.5ex] &= n(\mathrm{Prevacid}) ~ r(\mathrm{F,Prevacid}) ~ M(\mathrm{F}) \\[1.5ex] &= 0.8\bar{5}~\mathrm{mol~Prevacid} \left ( \dfrac{3~\mathrm{mol~F}}{\mathrm{mol~Prevacid}} \right ) \left ( \dfrac{19.0\bar{0}~\mathrm{g}}{\mathrm{mol~F}} \right ) \\[1.5ex] &= 4\bar{8}.85~\mathrm{g} \\[1.5ex] &= 49~\mathrm{g} \end{align*} \]

C. Number of C atoms in 0.85 mol Prevacid

\[ \begin{align*} N(\mathrm{C}) &= n(\mathrm{C}) ~ N_{\mathrm{A}} \\[1.5ex] &= n(\mathrm{Prevacid}) ~ r(\textrm{C,Prevacid}) ~ N_{\mathrm{A}} \\[1.5ex] &= 0.8\bar{5}~\mathrm{mol~Prevacid} ~ \left ( \dfrac{16~\mathrm{mol~C}}{\mathrm{mol~Prevacid}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}~\mathrm{C}}{\mathrm{mol}} \right ) \\[1.5ex] &= 8.\bar{1}89\times 10^{24} \\[1.5ex] &= 8.2\times 10^{24}~\mathrm{C}\\[3ex] \end{align*} \]

D. Mass (in g) of 5.25 × 10²¹ molecules of Prevacid

\[ \begin{align*} m(\mathrm{Prevacid}) &= n(\mathrm{Prevacid}) ~ M(\mathrm{Prevacid}) \\[1.5ex] &= N(\mathrm{Prevacid}) ~ N_{\mathrm{A}}^{-1} ~ M(\mathrm{Prevacid}) \\[1.5ex] &= 5.2\bar{5}\times 10^{21}~\mathrm{Prevacid} ~ \left ( \dfrac{\mathrm{mol}}{6.02\bar{2}\times 10^{23}~\mathrm{Prevacid}} \right ) \left ( \dfrac{369.3\bar{9}~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 3.2\bar{2}03~\mathrm{g~Prevacid} \\[1.5ex] &= 3.22~\mathrm{g~Prevacid} \end{align*} \]

Percent Composition, Percent Abundance

Guanidine, HNC(NH₂)₂, is a fertilizer. What is the percent by mass of nitrogen in the fertilizer?

Solution

Answer: 71.1 %

Concept: percent composition by mass

\[ \begin{align*} w(\mathrm{N})~\% &= w(\mathrm{N}) \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{3~ A_{\mathrm{r}}{^{\circ}}(\mathrm{N})} {M_{\mathrm{r}}(\mathrm{HNC(NH_2)_2})} \Biggl\} ~ \times ~ 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{3~A_{\mathrm{r}}{^{\circ}}(\mathrm{N})} { \left [ 5~A_{\mathrm{r}}{^{\circ}}(\mathrm{H}) + 3~A_{\mathrm{r}}{^{\circ}}(\mathrm{N}) + A_{\mathrm{r}}{^{\circ}}(\mathrm{C}) \right ] } \Biggl\} ~ \times ~ 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{3 \left(14.0\bar{1} \right)} { \left [ \left ( 5\times 1.0\bar{1} \right ) + \left ( 3\times 14.0\bar{1} \right ) + 12.0\bar{1} \right ] } \Biggl\} ~ \times ~ 100~\% \\[1.5ex] &= \left ( \dfrac{42.0\bar{3}}{59.\bar{0}9} \right ) \times 100~\% \\[1.5ex] &= 71.\bar{1}28~\% \\[1.5ex] &= 71.1~\% \end{align*} \]

Determine the percent by mass of Mg in chlorophyll (C₅₅H₇₂MgN₄O₅), the green pigment in plant cells.

Solution

Answer: 2.720 %

Concept: percent composition by mass

\[ \begin{align*} w(\mathrm{Mg})~\% &= w(\mathrm{Mg}) \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{A_{\mathrm{r}}{^{\circ}}(\mathrm{Mg})} {M_{\mathrm{r}}(\mathrm{chlorophyll})} \Biggl\} ~ \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{A_{\mathrm{r}}{^{\circ}}(\mathrm{Mg})} { \left [ 55 ~ A_{\mathrm{r}}{^{\circ}}(\mathrm{C}) + 72 ~ A_{\mathrm{r}}{^{\circ}}(\mathrm{H}) + A_{\mathrm{r}}{^{\circ}}(\mathrm{Mg}) + 4 ~ A_{\mathrm{r}}{^{\circ}}(\mathrm{N}) + 5 ~ A_{\mathrm{r}}{^{\circ}}(\mathrm{O}) \right ] } \Biggl\} ~ \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{ 24.3\bar{1} } { \left [ \left ( 55\times 12.0\bar{1} \right ) + \left ( 72\times 1.0\bar{1} \right ) + 24.3\bar{1} + \left ( 4\times 14.0\bar{1} \right ) + \left ( 5\times 16.0\bar{0} \right ) \right ] } \Biggl\} ~ \times 100~\% \\[1.5ex] &= \Biggl\{ \dfrac{24.3\bar{1}} {893.6\bar{2}} \Biggl\} ~ \times 100~\% \\[1.5ex] &= 2.72\bar{0}39~\% \\[1.5ex] &= 2.720~\% \end{align*} \]

The mineral spodumene has the formula LiAlSi₂O₆. What is the mass (in g) of lithium in a 438 g sample?

Solution

Answer: 16.3 g

Concept: mass fraction

We can get the mass of an element in a sample with a known mass by taking the mass fraction of the element and multiplying by the mass of the sample.

\[ \begin{align*} m(\mathrm{Li}) &= w(\mathrm{Li}) ~ m(\mathrm{sample}) \\[1.5ex] &= \left ( \dfrac{A_{\mathrm{r}}{^{\circ}}(\mathrm{Li})} {M_{\mathrm{r}}(\mathrm{LiAlSi_2O_6})} \right ) ~ m(\mathrm{sample}) \\[1.5ex] &= \left ( \dfrac{A_{\mathrm{r}}{^{\circ}}(\mathrm{Li})} { \left [ A_{\mathrm{r}}{^{\circ}}(\mathrm{Li}) + A_{\mathrm{r}}{^{\circ}}(\mathrm{Al}) + 2~A_{\mathrm{r}}{^{\circ}}(\mathrm{Si}) + 6~A_{\mathrm{r}}{^{\circ}}(\mathrm{O}) \right ] } \right ) ~ m(\mathrm{sample}) \\[1.5ex] &= \left ( \dfrac{6.9\bar{4}} { \left [ 6.9\bar{4} + 26.9\bar{8} + (2 \times 28.0\bar{9}) + (6 \times 16.0\bar{0}) \right ] } \right ) ~ \left ( 43\bar{8}~\mathrm{g} \right ) \\[1.5ex] &= \left ( \dfrac{6.9\bar{4}}{186.1\bar{0}}\right ) \left ( 43\bar{8}~\mathrm{g} \right ) \\[1.5ex] &= 16.\bar{3}33~\mathrm{g}\\[1.5ex] &= 16.3~\mathrm{g} \end{align*} \]

The mineral spodumene has the formula LiAlSi₂O₆. How many lithium atoms are present in a 101 g sample?

Solution

Answer: 3.27 × 10²³

Concept: moles; the Avogadro constant

\[ \begin{align*} N(\mathrm{Li}) &= n(\mathrm{Li})~ N_{\mathrm{A}} \\[1.5ex] &= m(\mathrm{Li}) ~ M(\mathrm{Li})^{-1} ~ N_{\mathrm{A}} \\[1.5ex] &= w(\mathrm{Li}) ~ m(\mathrm{sample}) ~ M(\mathrm{Li})^{-1} ~ N_{\mathrm{A}} \\[1.5ex] &= \left ( \dfrac{A_{\mathrm{r}}{^{\circ}}(\mathrm{Li})} {M_{\mathrm{r}}(\mathrm{LiAlSi_2O_6})} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{Li})^{-1} ~ N_{\mathrm{A}} \\[1.5ex] &= \left ( \dfrac{A_{\mathrm{r}}{^{\circ}}(\mathrm{Li})} { \left [ A_{\mathrm{r}}{^{\circ}}(\mathrm{Li}) + A_{\mathrm{r}}{^{\circ}}(\mathrm{Al}) + \left ( 2~A_{\mathrm{r}}{^{\circ}}(\mathrm{Si}) \right ) + \left ( 6~A_{\mathrm{r}}{^{\circ}}(\mathrm{O}) \right ) \right ] } \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{Li})^{-1} ~ N_{\mathrm{A}} \\[1.5ex] &= \left ( \dfrac{6.94} { \left [ 6.9\bar{4} + 26.9\bar{8} + \left ( 2 \times 28.0\bar{9} \right ) + \left ( 6 \times 16.0\bar{0} \right ) \right ] } \right ) ~ \left ( \dfrac{10\bar{1}~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{6.9\bar{4}~\mathrm{g}}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= \left ( \dfrac{6.9\bar{4}}{186.1\bar{0}}\right ) \left ( \dfrac{10\bar{1}~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{6.9\bar{4}~\mathrm{g}}} \right ) \left ( \dfrac{6.02\bar{2}\times 10^{23}}{\mathrm{mol}} \right ) \\[1.5ex] &= 3.2\bar{6}82\times 10^{23} \\[1.5ex] &= 3.27\times 10^{23} \end{align*} \]

Find the mass (in g as normalized scientific notation) of 100. atoms of iron.

Solution

Answer: 9.27 × 10^–21 g

Concept: moles; the Avogadro constant

\[ \begin{align*} m(\mathrm{Fe}) &= n(\mathrm{Fe}) ~ M(\mathrm{Fe}) \\[1.5ex] &= N(\mathrm{Fe}) ~ N_{\mathrm{A}}{^{-1}} ~ M(\mathrm{Fe}) \\[1.5ex] &= 10\bar{0}.~\mathrm{Fe} ~ \left ( \dfrac{\mathrm{mol}}{6.02\bar{2}\times 10^{23}~\mathrm{Fe}} \right ) \left ( \dfrac{55.8\bar{5}~\mathrm{g}}{\mathrm{mol}} \right ) \\[1.5ex] &= 9.2\bar{7}43 \times 10^{-21}~\mathrm{g} \\[1.5ex] &= 9.27 \times 10^{-21}~\mathrm{g} \end{align*} \]

Analysis of a sample of a covalent compound showed that it contained 14.4 % hydrogen and 85.6 % carbon by mass. What is the empirical formula for the compound?

Solution

Answer: CH₂

Concept: percent composition by mass; empirical formula

The empirical formula represents the simplest whole-number ratio of moles of each element in a compound. We will follow a three-step process to find this ratio from the given mass percentages.

NoteAssuming a 100 g Sample

Because percent composition is a ratio, it is independent of the total mass of the sample. To simplify the calculation, we can assume a convenient total mass. Assuming an exactly 100 g sample is the standard and most direct method, as it allows us to convert the percentages directly into grams (e.g., 14.4% of 100 g is exactly 14.4 g).

Find the moles of H in the sample:

\[ \begin{align*} w(\mathrm{H})~\% &= w(\mathrm{H}) \times 100~\% \\[1.5ex] &= m(\mathrm{H})~ m(\mathrm{sample})^{-1}~ \times 100~\% \\[1.5ex] &= n(\mathrm{H}) ~ M(\mathrm{H})~ m(\mathrm{sample})^{-1}~ \times 100~\% \longrightarrow \\[1.5ex] n(\mathrm{H}) &= \left ( \dfrac{w\left(\mathrm{H}\right)\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{H})^{-1} \\[1.5ex] &= \left ( \dfrac{14.\bar{4}~\%}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{1.0\bar{1}~g}} \right ) \\[1.5ex] &= 14.\bar{2}57~\mathrm{mol~H} \end{align*} \]

Find the moles of C in the sample:

\[ \begin{align*} w(\mathrm{C})~\% &= w(\mathrm{C}) \times 100~\% \\[1.5ex] &= m(\mathrm{C})~ m(\mathrm{sample})^{-1}~ \times 100~\% \\[1.5ex] &= n(\mathrm{C}) ~ M(\mathrm{C})~ m(\mathrm{sample})^{-1}~ \times 100~\% \longrightarrow \\[1.5ex] n(\mathrm{C}) &= w(\mathrm{C}) ~ m(\mathrm{sample}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \left ( \dfrac{w\left(\mathrm{C}\right)\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \left ( \dfrac{85.\bar{6}~\%}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{12.0\bar{1}~g}} \right ) \\[1.5ex] &= 7.1\bar{2}7~\mathrm{mol~C} \end{align*} \]

Find the mole ratio of H–to–C in the sample:

\[ \begin{align*} r(\mathrm{H,C}) &= n(\mathrm{H})~ n(\mathrm{C})^{-1}\\[1.5ex] &= \dfrac{14.\bar{2}57~\mathrm{mol~H}}{7.1\bar{2}7~\mathrm{mol~C}}\\[1.5ex] &= \dfrac{2.00~\mathrm{mol~H}}{1~\mathrm{mol~C}} \\[2ex] &\mathrm{CH_2} \end{align*} \]

Find the percent composition by mass of each element in Yttrium barium copper oxide (YBa₂Cu₃O₇), a high-temperature superconductor compound.

Solution

Answer: Y: 13.34 %; Ba: 41.23 %; Cu: 28.62 %; O: 16.81 %

Concept: percent composition by mass

Find the relative molecular mass of YBa₂Cu₃O₇.

\[ \begin{align*} M_{\mathrm{r}}(\mathrm{YBa_2Cu_3O_7}) &= A_{\mathrm{r}}{^{\circ}}(\mathrm{Y}) + \left ( 2~A_{\mathrm{r}}{^{\circ}}(\mathrm{Ba}) \right ) + \left ( 3~A_{\mathrm{r}}{^{\circ}}(\mathrm{Cu}) \right ) + \left ( 7~A_{\mathrm{r}}{^{\circ}}(\mathrm{O}) \right ) \\[1.5ex] &= 88.9\bar{1} + \left ( 2 \times 137.3\bar{3} \right ) + \left ( 3 \times 63.5\bar{5} \right ) + \left ( 7 \times 16.0\bar{0} \right ) \\[1.5ex] &= 666.2\bar{2}0 \\[1.5ex] &= 666.22 \end{align*} \]

Find the mass percent of each element in YBa₂Cu₃O₇.

$$ \[\begin{align*} w(\mathrm{Y})~\% &= w(\mathrm{Y}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{A_{\mathrm{r}}{^{\circ}}(\mathrm{Y})}{M_{\mathrm{r}}(\mathrm{YBa_2Cu_3O_7})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{88.9\bar{1}}{666.2\bar{2}0} \right ) \times 100~\% \\[1.5ex] &= 13.3\bar{4}54~\% \\[1.5ex] &= 13.34~\%~\mathrm{Y} \\[3ex] w(\mathrm{Ba})~\% &= w(\mathrm{Ba}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2~A_{\mathrm{r}}{^{\circ}}(\mathrm{Ba})}{M_{\mathrm{r}}(\mathrm{YBa_2Cu_3O_7})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{2\times 137.3\bar{3}}{666.2\bar{2}0} \right ) \times 100~\% \\[1.5ex] &= 41.2\bar{2}66~\% \\[1.5ex] &= 41.23~\%~\mathrm{Ba} \\[3ex] w(\mathrm{Cu})~\% &= w(\mathrm{Cu}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{3~A_{\mathrm{r}}{^{\circ}}(\mathrm{Cu})}{M_{\mathrm{r}}(\mathrm{YBa_2Cu_3O_7})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{3\times 63.5\bar{5}}{666.2\bar{2}0} \right ) \times 100~\% \\[1.5ex] &= 28.6\bar{1}66~\% \\[1.5ex] &= 28.62~\%~\mathrm{Cu} \\[3ex] w(\mathrm{O})~\% &= w(\mathrm{O}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{7~A_{\mathrm{r}}{^{\circ}}(\mathrm{O})}{M_{\mathrm{r}}(\mathrm{YBa_2Cu_3O_7})} \right ) \times 100~\% \\[1.5ex] &= \left ( \dfrac{7\times 16.0\bar{0}}{666.2\bar{2}0} \right ) \times 100~\% \\[1.5ex] &= 16.8\bar{1}12~\% \\[1.5ex] &= 16.81~\%~\mathrm{O} \end{align*}\] $$

Hemoglobin is a protein that transports oxygen in mammals. Hemoglobin is 0.347 % Fe (by mass). Each hemoglobin molecule contains 4 Fe atoms. What is the molar mass (in g mol^–1 as normalized scientific notation) of hemoglobin?

Solution

Answer: 6.44 × 10⁴ g mol^–1

Concept: percent composition by mass

\[ \begin{align*} w(\mathrm{Fe})~\% &= w(\mathrm{Fe}) \times 100~\% \\[1.5ex] &= \left ( \dfrac{4~A_{\mathrm{r}}{^{\circ}}(\mathrm{Fe})}{M_{\mathrm{r}}(\mathrm{hemoglobin})} \right ) \times 100~\% \longrightarrow \\[1.5ex] M_{\mathrm{r}}(\mathrm{hemoglobin}) &= \left ( \dfrac{4~A_{\mathrm{r}}{^{\circ}}(\mathrm{Fe})} {w(\mathrm{Fe})~\%} \right ) \times 100~\% \\[1.5ex] M(\mathrm{hemoglobin}) &= \left ( \dfrac{4~A_{\mathrm{r}}{^{\circ}}(\mathrm{Fe})} {w(\mathrm{Fe})~\%} \right ) ~ M_{\mathrm{u}} \times 100~\% \\[1.5ex] &= \left ( \dfrac{4~(55.8\bar{5})} {0.34\bar{7}~\%} \right ) \left ( \dfrac{\mathrm{g}}{\mathrm{mol}} \right ) \times 100~\% \\[1.5ex] &= 6.4\bar{3}80 \times 10^{4}~\mathrm{g~mol^{-1}} \\[1.5ex] &= 6.44 \times 10^{4}~\mathrm{g~mol^{-1}} \end{align*} \]

A compound that only contains carbon, hydrogen, and oxygen is 48.64 % C and 8.16 % H (by mass). What is the empirical formula of this substance?

Solution

Answer: C₃H₆O₂

Concept: percent composition by mass; empirical formula

Assume an exact 100 g sample.

Find the moles of C, H, and O in an exact 100 g sample.

\[ \begin{align*} n(\mathrm{C}) &= m(\mathrm{C}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= w(\mathrm{C}) ~ m(\mathrm{sample}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \left ( \dfrac{w(\mathrm{C})~\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{C})^{-1} \\[1.5ex] &= \left ( \dfrac{48.6\bar{4}~\%~\mathrm{C}}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{12.0\bar{1}~g}} \right ) \\[1.5ex] &= 4.04\bar{9}95~\mathrm{mol~C}\\[3ex] n(\mathrm{H}) &= m(\mathrm{H}) ~ M(\mathrm{H})^{-1} \\[1.5ex] &= w(\mathrm{H}) ~ m(\mathrm{sample}) ~ M(\mathrm{H})^{-1} \\[1.5ex] &= \left ( \dfrac{w(\mathrm{H})~\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{H})^{-1} \\[1.5ex] &= \left ( \dfrac{8.1\bar{6}~\%~\mathrm{H}}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{1.0\bar{1}~g}} \right ) \\[1.5ex] &= 8.07\bar{9}20~\mathrm{mol~H}\\[3ex] n(\mathrm{O}) &= m(\mathrm{O}) ~ M(\mathrm{O})^{-1} \\[1.5ex] &= w(\mathrm{O}) ~ m(\mathrm{sample}) ~ M(\mathrm{O})^{-1} \\[1.5ex] &= \left ( \dfrac{w(\mathrm{O})~\%}{100~\%} \right ) ~ m(\mathrm{sample}) ~ M(\mathrm{O})^{-1} \\[1.5ex] &= \left ( \dfrac{\left [ 100 - (48.6\bar{4} + 8.1\bar{6}) \right ]~\%~\mathrm{O}}{100~\%}\right ) \left ( \dfrac{100~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{\mathrm{16.0\bar{0}~g}} \right ) \\[1.5ex] &= 2.70\bar{0}00~\mathrm{mol~O}\\[3ex] \end{align*} \]

Find the mole ratio of C, H, and O to a smallest whole number ratio by dividing by the smallest mole number and multiplying by an integer (here the integer is 2).

\[ \begin{equation*} \begin{gathered} \mathrm{C}_{4.04\bar{9}95}\mathrm{H}_{8.07\bar{9}20}\mathrm{O}_{2.70\bar{0}00} \\[1.5ex] \mathrm{C}_{\frac{4.04\bar{9}9}{2.70\bar{0}00}}\mathrm{H}_{\frac{8.07\bar{9}2}{2.70\bar{0}00}}\mathrm{O}_{\frac{{2. 70\bar{0}00}}{2.70\bar{0}00}} \\[1.5ex] \mathrm{C}_{1.5}\mathrm{H}_{3}\mathrm{O}_{} \\[1.5ex] \mathrm{C}_{1.5\times 2}\mathrm{H}_{3\times 2}\mathrm{O}_{1\times 2} \\[4ex] \mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}_{2} \end{gathered} \end{equation*} \]