Question

for questions 29 - 30, consider a compound consisting of 55.80% c, 7.04% h, and 37.17% o by mass. 29. what is the empirical formula for the compound? a) ch₂o b) c₂h₃o c) c₂h₆o d) c₄h₇o₂ 30. the formula weight for the compound is 129.15 amu. what is the molecular formula for the compound? a) c₂h₃o b) c₆h₉o₃ c) c₅h₁₁o₄ d) c₆h₁₂o₆

Explanation:

Step1: Assume 100g of the compound

Since we have mass - percentages, assuming 100g of the compound means we have 55.80g of C, 7.04g of H, and 37.17g of O.

Step2: Calculate the number of moles of each element

The molar - mass of C is \(M_{C}=12.01g/mol\), the molar - mass of H is \(M_{H}=1.01g/mol\), and the molar - mass of O is \(M_{O}=16.00g/mol\).
The number of moles of C, \(n_{C}=\frac{55.80g}{12.01g/mol}\approx4.65mol\).
The number of moles of H, \(n_{H}=\frac{7.04g}{1.01g/mol}\approx6.97mol\).
The number of moles of O, \(n_{O}=\frac{37.17g}{16.00g/mol}\approx2.32mol\).

Step3: Find the mole - ratio of the elements

Divide each number of moles by the smallest number of moles (\(n_{O} = 2.32mol\)).
\(\frac{n_{C}}{n_{O}}=\frac{4.65mol}{2.32mol}\approx2\), \(\frac{n_{H}}{n_{O}}=\frac{6.97mol}{2.32mol}\approx3\), \(\frac{n_{O}}{n_{O}} = 1\).
So the empirical formula is \(C_{2}H_{3}O\).

Step4: Calculate the empirical - formula weight

The empirical - formula weight of \(C_{2}H_{3}O\) is \(M_{empirical}=2\times12.01g/mol + 3\times1.01g/mol+16.00g/mol=24.02g/mol + 3.03g/mol+16.00g/mol = 43.05g/mol\).

Step5: Find the multiple \(n\) to get the molecular formula

We know that the formula weight of the compound is 129.15 amu (g/mol).
\(n=\frac{M_{molecular}}{M_{empirical}}=\frac{129.15g/mol}{43.05g/mol}=3\).
Multiply the sub - scripts in the empirical formula \(C_{2}H_{3}O\) by 3 to get the molecular formula \(C_{6}H_{9}O_{3}\).

Answer:

1. B. \(C_{2}H_{3}O\)
2. B. \(C_{6}H_{9}O_{3}\)