Question

application questions

1. when 0.519 g of pure sulfur (s) is allowed to react with fluorine gas, a colorless corrosive gas forms. if the product weighs 0.642 g, what is the empirical formula of the gaseous product? show your work.
2. the elemental composition of an unknown compound was found to be 74.0% carbon, 8.7% hydrogen and 17.3% nitrogen by mass.

(a) what is the empirical formula of the compound?
(b) the molar mass of the compound was determined to be 162.23 g/mol. what is the molecular formula of the compound?

Explanation:

Step1: Find mass of fluorine in product

The mass of fluorine ($m_F$) in the product is the mass of the product minus the mass of sulfur.
$m_F=0.642 - 0.519=0.123$ g

Step2: Calculate moles of sulfur and fluorine

The molar - mass of sulfur ($S$) is $M_S = 32.06$ g/mol, and the molar - mass of fluorine ($F$) is $M_F=18.998$ g/mol.
The number of moles of sulfur ($n_S$) is $n_S=\frac{m_S}{M_S}=\frac{0.519}{32.06}\approx0.0162$ mol
The number of moles of fluorine ($n_F$) is $n_F=\frac{m_F}{M_F}=\frac{0.123}{18.998}\approx0.00647$ mol

Step3: Find mole ratio

Divide each number of moles by the smaller number of moles to get the mole ratio.
For sulfur: $\frac{n_S}{n_F}=\frac{0.0162}{0.00647}\approx2.5$
For fluorine: $\frac{n_F}{n_F} = 1$
Multiply by 2 to get whole - number ratios. The ratio of $S:F$ is $5:2$

for 2(a):

Step1: Assume 100 g of the compound

If we assume 100 g of the compound, then the masses of carbon ($m_C$), hydrogen ($m_H$), and nitrogen ($m_N$) are 74.0 g, 8.7 g, and 17.3 g respectively.

Step2: Calculate moles of each element

The molar - mass of carbon $M_C = 12.01$ g/mol, hydrogen $M_H=1.008$ g/mol, and nitrogen $M_N = 14.01$ g/mol.
The number of moles of carbon $n_C=\frac{m_C}{M_C}=\frac{74.0}{12.01}\approx6.16$ mol
The number of moles of hydrogen $n_H=\frac{m_H}{M_H}=\frac{8.7}{1.008}\approx8.63$ mol
The number of moles of nitrogen $n_N=\frac{m_N}{M_N}=\frac{17.3}{14.01}\approx1.235$ mol

Step3: Find mole ratio

Divide each number of moles by the smallest number of moles ($n_N$).
For carbon: $\frac{n_C}{n_N}=\frac{6.16}{1.235}\approx5$
For hydrogen: $\frac{n_H}{n_N}=\frac{8.63}{1.235}\approx7$
For nitrogen: $\frac{n_N}{n_N}=1$

for 2(b):

Step1: Calculate the empirical - formula mass

The empirical - formula mass ($M_{EF}$) of $C_5H_7N$ is $M_{EF}=5\times12.01+7\times1.008 + 1\times14.01=60.05+7.056+14.01 = 81.116$ g/mol

Step2: Find the multiple ($n$)

The multiple $n$ is given by $n=\frac{M}{M_{EF}}$, where $M = 162.23$ g/mol is the molar mass of the compound.
$n=\frac{162.23}{81.116}\approx2$

Step3: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by $n$.
The molecular formula is $(C_5H_7N)_2=C_{10}H_{14}N_2$

Answer:

The empirical formula of the gaseous product is $S_5F_2$