Percentage composition | Empirical formula and Molecular formula | Worksheet

QUESTION 1:

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

SOLUTION:

$\%\text{C}=\frac{7.34\text{g C}}{12.04\text{g compound}}\times 100\%=61.0\%$

$\%\text{H}=\frac{1.85\text{g H}}{12.04\text{g compound}}\times 100\%=15.4\%$

$\%\text{N}=\frac{2.85\text{g N}}{12.04\text{g compound}}\times 100\%=23.7\%$

QUESTION 2: 

Aspirin is a compound with the molecular formula C9H8O4. What is its percent composition?

SOLUTION: 

$\begin{array}{cc}\%\text{C =}& \frac{9\text{mol C}\times \text{molar mass C}}{\text{molar mass}{\text{C}}_9{\text{H}}_{18}{\text{O}}_4}\times 100=\frac{9\times 12.01\text{g/mol}}{180.159\text{g/mol}}\times 100=\frac{108.09\text{g/mol}}{180.159\text{g/mol}}\times 100\\ \%\text{C}& 60.00\%\text{C}\end{array}$

$\begin{array}{cc}\%\text{H =}& \frac{8\text{mol H}\times \text{molar mass H}}{\text{molar mass}{\text{C}}_9{\text{H}}_{18}{\text{O}}_4}\times 100=\frac{8\times 1.008\text{g/mol}}{180.159\text{g/mol}}\times 100=\frac{8.064\text{g/mol}}{180.159\text{g/mol}}\times 100\\ \%\text{H}& 4.476\%\text{H}\end{array}$

$\begin{array}{cc}\%\text{O =}& \frac{4\text{mol O}\times \text{molar mass O}}{\text{molar mass}{\text{C}}_9{\text{H}}_{18}{\text{O}}_4}\times 100=\frac{4\times 16.00\text{g/mol}}{180.159\text{g/mol}}\times 100=\frac{64.00\text{g/mol}}{180.159\text{g/mol}}\times 100\\ \%\text{O}& 35.52\%\text{O}\end{array}$

QUESTION 3: 

A sample of hematite contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?

SOLUTION: 

Step-1: 

$\begin{array}{cc}34.97\text{g Fe}\left(\frac{\text{mol Fe}}{55.85\text{g}}\right)& 0.6261\text{mol Fe}\\ 15.03\text{g O}\left(\frac{\text{mol O}}{16.00\text{g}}\right)& 0.9394\text{mol O}\end{array}$

Step-2: 

$\begin{array}{cc}\frac{0.6261}{0.6261}& \mathrm{=\; 1.000}\text{mol Fe}\\ \frac{0.9394}{0.6261}& \mathrm{=\; 1.500}\text{mol O}\end{array}$

Step 3:     

2 (Fe1O1.5)=Fe2O3

The empirical formula is Fe2O3

QUESTION 4: 

The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O. What is the empirical formula for this gas?

SOLUTION: 

Step-1: 

$\begin{array}{cc}27.29\%\text{C}& \frac{27.29\text{g C}}{100\text{g compound}}\\ 72.71\%\text{O}& \frac{72.71\text{g O}}{100\text{g compound}}\end{array}$

Step-2: 

$\begin{array}{cc}27.29\%\text{C}\left(\frac{\text{mol C}}{12.01\text{g}}\right)& 2.272\text{mol C}\\ 72.71\%\text{O}\left(\frac{\text{mol O}}{16.00\text{g}}\right)& 4.544\text{mol O}\end{array}$

Step 3: 

$\begin{array}{cc}\frac{2.272\text{g C}}{2.272}& \mathrm{=\; 1}\\ \frac{4.544\text{g O}}{2.272}& \mathrm{=\; 2}\end{array}$

So, the empirical formula is CO2.

QUESTION 5: 

Nicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains 74.02% C, 8.710% H, and 17.27% N. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?

SOLUTION: 

Step-1: 

$\begin{array}{cc}\left(74.02\text{g C}\right)\left(\frac{1\text{mol C}}{12.01\text{g C}}\right)& 6.163\text{mol C}\\ \left(8.710\text{g H}\right)\left(\frac{1\text{mol H}}{1.01\text{g H}}\right)& 8.624\text{mol H}\\ \left(17.27\text{g N}\right)\left(\frac{1\text{mol N}}{14.01\text{g N}}\right)& 1.233\text{mol N}\end{array}$

Step-2: 

$\begin{array}{cc}6.163\text{mol C/}1.233\text{mol N}& \mathrm{=\; 5}\\ 8.264\text{mol H/}1.233\text{mol N}& \mathrm{=\; 7}\\ 1.233\text{mol N/}1.233\text{mol N}& \mathrm{=\; 1}\end{array}$

Step-3:

$\begin{array}{cc}\frac{1.233}{1.233}& \mathrm{=\; 1.000}\text{mol N}\\ \frac{6.163}{1.233}& \mathrm{=\; 4.998}\text{mol C}\\ \frac{8.624}{1.233}& \mathrm{=\; 6.994}\text{mol H}\end{array}$

The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C5H7N

We calculate the molar mass for nicotine from the given mass and molar amount of compound:

$\frac{40.57\text{g nicotine}}{0.2500\text{mol nicotine}}=\frac{162.3\text{g}}{\text{mol}}$

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

$\frac{162.3\text{g/mol}}{81.13\frac{\text{g}}{\text{formula unit}}}=2\text{formula units/molecule}$

Thus, we can derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

$({\text{C}}_5{\text{H}}_7\text{N}{)}_2={\text{C}}_{10}{\text{H}}_{14}{\text{N}}_2$

QUESTION 6: 

What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.48% O, and a molecular mass of 194.2 amu?

QUESTION 7: 

Determine the empirical formulas for compounds with the following percent compositions:

(a) 43.6% phosphorus and 56.4% oxygen

(b) 28.7% K, 1.5% H, 22.8% P, and 47.0% O

QUESTION 8: 

A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula?

QUESTION 9: 

A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.

QUESTION 10: 

Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol.

QUESTION 11: 

Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula?

QUESTION 12: 

Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers:

(a) Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O

(b) Saran; 24.8% C, 2.0% H, 73.1% Cl

(c) polyethylene; 86% C, 14% H

(d) polystyrene; 92.3% C, 7.7% H

(e) Orlon; 67.9% C, 5.70% H, 26.4% N