Question

calculate the percent composition by mass of oxygen in the fictional compound below. include the units, but do not write the chemical formula. round the answer to 3 significant figures. nh₄(c₃h₄o₃)₂ oxygen =

Explanation:

Step1: Calculate molar mass of compound

First, find the number of each atom:

• N: 1
• H: \(4 + 4\times2 = 12\) (4 from \(NH_4\), 4×2 from \((C_3H_4O_3)_2\))
• C: \(3\times2 = 6\)
• O: \(3\times2 = 6\)

Molar masses (g/mol): N = 14.01, H = 1.008, C = 12.01, O = 16.00.

Molar mass of \(NH_4(C_3H_4O_3)_2\):
\(1\times14.01 + 12\times1.008 + 6\times12.01 + 6\times16.00\)
\(= 14.01 + 12.096 + 72.06 + 96.00\)
\(= 14.01 + 12.096 = 26.106\); \(26.106 + 72.06 = 98.166\); \(98.166 + 96.00 = 194.166\) g/mol.

Step2: Calculate mass of O

Mass of O: \(6\times16.00 = 96.00\) g/mol.

Step3: Calculate percent composition

Percent composition of O: \(\frac{\text{Mass of O}}{\text{Molar mass of compound}} \times 100\)
\(\frac{96.00}{194.166} \times 100 \approx 49.4\%\) (rounded to 3 significant figures).

Answer:

49.4%