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`

Step-by-step 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}}= (0.1605g)/(12g/mole)=0.0134moles`

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

Moles of S =
`\frac{\text{ given mass of S}}{\text{ molar mass of S}}= (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 =
`(0.0134)/(0.0044)=3`

For H =
`(0.0220)/(0.0044)=5`

For S =
`(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}}=(146)/(73)=2`

The molecular formula will be=
`2* C_3H_5S=C_6H_(10)S_2`