Question

A compound has a molecular weight of 146 g/mol. A 0.3250 g sample of the compound contains 0.1605 g of carbon, 0.0220 g of hydrogen, and 0.1425 g of sulfur. What is the molecular formula of the compound

Answer 1

Answer: The molecular formula is $C_6H_{10}S_2$

Explanation:

We are given:

Mass of $C$ = 0.1605 g

Mass of $H$= 0.0220 g

mass of $S$ = 0.1425 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{0.1605g}{12g/mole}=0.0134moles$

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.0220g}{1g/mole}=0.0220moles$

Moles of S =$\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{0.1425g}{32g/mole}=0.0044moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.0134}{0.0044}=3$

For H = $\frac{0.0220}{0.0044}=5$

For S =$\frac{0.0044}{0.0044}=1$

The ratio of C : H: S=  3: 5: 1

Hence the empirical formula is $C_3H_5S$

The empirical weight of $C_3H_5S$ = 3(12)+5(1)+1(32)= 73g.

The molecular weight = 146 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{146}{73}=2$

The molecular formula will be=$2\times C_3H_5S=C_6H_{10}S_2$