Question

A compound responsible for the odor of garlic has a molecular weight of 146 g/mol. A 0.650 g sample of the compound contains 0.321 g of carbon, 0.044 g of hydrogen, and 0.285 g of sulfur. What is the molecular formula of the compound?

Answer 1

`molar mass =146g/mol\\</p><p>amount of carbon=0.321g\\</p><p>moler of carbon=(0.321)/(12.01)=0.026----(1)\\</p><p>amount of H=0.044g=0.043--------(2)\\</p><p>mole of H=(0.044)/(1.01)\\</p><p>amount of surface=0.285\\</p><p>mole of (s)=(0.285)/(32.06)=0.0088--------(3)\\</p><p>dividing by the smallest unit is (1),(2),(3)\\</p><p>(0.026)/(0.088)=2.95\approx 3\\</p><p>(0.043)/(0.0088)=4.88\approx 5\\</p><p>(0.0088)/(0.0088)=1\\`
`Empirical formula of the compound is c_(3)H_(5)S\\</p><p>at the let of C_(3)H_(5)S=73\\</p><p>molecular weight=146\\</p><p>molecularity=(146)/(73)=2\\</p><p>so the formula of the compound is c_(6)H_(10)S_(2)`

Answer 2

Answer : The molecular formula of a compound is,
`C_6H_(10)S_2`

Solution : Given,

Mass of C = 0.321 g

Mass of H = 0.044 g

Mass of S = 0.285 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of S = 32 g/mole

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (0.321g)/(12g/mole)=0.0267moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (0.044g)/(1g/mole)=0.044moles`

Moles of S =
`\frac{\text{ given mass of S}}{\text{ molar mass of S}}= (0.285g)/(32g/mole)=0.0089moles`

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

For C =
`(0.0267)/(0.0089)=3`

For H =
`(0.044)/(0.0089)=4.94\approx 5`

For S =
`(0.0089)/(0.0089)=1`

The ratio of C : H : S = 3 : 5 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula =
`C_3H_5S_1`

The empirical formula weight = 3(12) + 5(1) + 1(32) = 73 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

`n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}`

`n=(146)/(73)=2`

Molecular formula =
`(C_3H_5S_1)_n=(C_3H_5S_1)_2=C_6H_(10)S_2`

Therefore, the molecular of the compound is,
`C_6H_(10)S_2`