Question

calculate the molar mass of each of the following compounds, then determine the percent composition of each element in the compound. round the atomic mass numbers and the percents to two decimal places.

1. what is the percent composition of carbon and oxygen in co₂?
2. what is the percent composition of carbon and hydrogen in c₂h₆?
3. what is the percent composition of iron in feo?
4. what is the percent composition of calcium in calcium phosphate ca₃(po₄)₂?

Explanation:

Step1: Find molar mass of CO

Atomic mass: C = 12.01 g/mol, O = 16.00 g/mol
Molar mass of CO = $12.01 + 16.00 = 28.01$ g/mol

Step2: % composition of C in CO

$\text{% C} = \frac{12.01}{28.01} \times 100 = 42.88\%$

Step3: % composition of O in CO

$\text{% O} = \frac{16.00}{28.01} \times 100 = 57.12\%$
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Step4: Find molar mass of $\text{C}_2\text{H}_6$

Atomic mass: C = 12.01 g/mol, H = 1.008 g/mol
Molar mass of $\text{C}_2\text{H}_6$ = $(2 \times 12.01) + (6 \times 1.008) = 24.02 + 6.048 = 30.068$ g/mol

Step5: % composition of C in $\text{C}_2\text{H}_6$

$\text{% C} = \frac{24.02}{30.068} \times 100 = 79.89\%$

Step6: % composition of H in $\text{C}_2\text{H}_6$

$\text{% H} = \frac{6.048}{30.068} \times 100 = 20.11\%$
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Step7: Find molar mass of FeO

Atomic mass: Fe = 55.85 g/mol, O = 16.00 g/mol
Molar mass of FeO = $55.85 + 16.00 = 71.85$ g/mol

Step8: % composition of Fe in FeO

$\text{% Fe} = \frac{55.85}{71.85} \times 100 = 77.73\%$
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Step9: Find molar mass of $\text{Ca}_3(\text{PO}_4)_2$

Atomic mass: Ca = 40.08 g/mol, P = 30.97 g/mol, O = 16.00 g/mol
Molar mass = $(3 \times 40.08) + (2 \times 30.97) + (8 \times 16.00) = 120.24 + 61.94 + 128.00 = 310.18$ g/mol

Step10: % composition of Ca in $\text{Ca}_3(\text{PO}_4)_2$

$\text{% Ca} = \frac{120.24}{310.18} \times 100 = 38.76\%$

Answer:

1. Percent composition of CO: Carbon = 42.88%, Oxygen = 57.12%
2. Percent composition of $\text{C}_2\text{H}_6$: Carbon = 79.89%, Hydrogen = 20.11%
3. Percent composition of Fe in FeO: 77.73%
4. Percent composition of Ca in $\text{Ca}_3(\text{PO}_4)_2$: 38.76%