Question

multiple choice 10 points: determine the mass percent of sulfur in caso₄
47%
58.3%
29.44%
23.56%
multiple choice 10 points: 16.0g of a hydrocarbon is combusted fully to form carbon dioxide and water. if the reaction produces 52.8g of carbon dioxide, determine the molecular weight of the empirical formula of the hydrocarbon. note that the hydrocarbon comprises h and c atoms only.
22.0 amu
14.0 amu
40.1 amu
28.0 amu

Explanation:

Step1: Calculate molar mass of CaSO₄

The molar mass of Ca = 40.08 g/mol, S = 32.07 g/mol, O = 16.00 g/mol.
$M_{CaSO_4}=40.08 + 32.07+4\times16.00=40.08 + 32.07 + 64.00=136.15$ g/mol

Step2: Calculate mass - percent of sulfur

The mass - percent of sulfur in CaSO₄ is $\frac{32.07}{136.15}\times100\%\approx23.56\%$

Step1: Calculate moles of carbon in CO₂

The molar mass of CO₂ is $M_{CO_2}=12.01 + 2\times16.00=44.01$ g/mol.
The moles of CO₂ produced is $n_{CO_2}=\frac{52.2}{44.01}\text{mol}\approx1.186$ mol.
Since 1 mole of CO₂ contains 1 mole of carbon, the moles of carbon in the hydrocarbon is also 1.186 mol.
The mass of carbon is $m_C = 1.186\times12.01\approx14.24$ g.

Step2: Calculate mass of hydrogen in the hydrocarbon

The mass of the hydrocarbon is 16.0 g, so the mass of hydrogen is $m_H=16.0 - 14.24 = 1.76$ g.
The moles of hydrogen is $n_H=\frac{1.76}{1.008}\text{mol}\approx1.75$ mol.

Step3: Determine the empirical - formula ratio

The ratio of C to H is $\frac{n_C}{n_H}=\frac{1.186}{1.75}\approx\frac{1}{1.5}=\frac{2}{3}$. So the empirical formula is C₂H₃.
The empirical - formula weight is $2\times12.01+3\times1.008 = 24.02+3.024 = 27.044\approx28.0$ amu

Answer:

D. 23.56%