Question

8
how many moles are in 1.00 g of propane (c3h8)? *
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9
determine the molecular formula of a compound with an empirical formula of nh2 and a formula mass of 32.06 amu. *
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10
when a 30.98 g sample of phosphorus reacts with oxygen, a 71.00 g sample of phosphorus oxide is formed. what is the percent of oxygen in this compound? round your answer to 2 decimal places. *

Explanation:

Question 8

Step1: Calculate molar mass of C₃H₈

Molar mass of C: \( 12.01 \, \text{g/mol} \), H: \( 1.008 \, \text{g/mol} \).
For \( \text{C}_3\text{H}_8 \), molar mass \( = 3\times12.01 + 8\times1.008 \)
\( = 36.03 + 8.064 = 44.094 \, \text{g/mol} \)

Step2: Use moles formula (\( n = \frac{m}{M} \))

Given \( m = 1.00 \, \text{g} \), \( M = 44.094 \, \text{g/mol} \)
\( n = \frac{1.00}{44.094} \approx 0.0227 \, \text{mol} \)

Step1: Calculate empirical formula mass of NH₂

Molar mass of N: \( 14.01 \, \text{g/mol} \), H: \( 1.008 \, \text{g/mol} \)
Empirical mass \( = 14.01 + 2\times1.008 = 14.01 + 2.016 = 16.026 \, \text{amu} \)

Step2: Find ratio (\( n = \frac{\text{Molecular mass}}{\text{Empirical mass}} \))

Given molecular mass \( = 32.06 \, \text{amu} \)
\( n = \frac{32.06}{16.026} \approx 2 \)

Step3: Multiply empirical formula by \( n \)

Molecular formula: \( (\text{NH}_2)_2 = \text{N}_2\text{H}_4 \)

Step1: Find mass of oxygen

Mass of compound (phosphorus oxide) \( = 71.00 \, \text{g} \)
Mass of phosphorus \( = 30.98 \, \text{g} \)
Mass of oxygen \( = 71.00 - 30.98 = 40.02 \, \text{g} \)

Step2: Calculate percent oxygen (\( \% \text{O} = \frac{\text{mass O}}{\text{mass compound}} \times 100 \))

\( \% \text{O} = \frac{40.02}{71.00} \times 100 \approx 56.37\% \)

Answer:

\( 0.0227 \) (or rounded as needed, e.g., 0.0227 mol)

Question 9