A compound is used as a food additive. The compound has a molar mass of 176.124 grams/mole. A 692.5-gram sample undergoes decomposition, producing 283.4 grams of carbon, 31.7 grams of hydrogen, and 377.4 grams of oxygen. What is the molecular formula of the compound?

Solution:- Grams of C, H and O are given from which the moles are calculated. Moles of each element are divided by the least one of them to get the mole ratio. If it's not a whole number ratio then we multiply the ratios by an smallest integer like 2,3,4------etc to get the whole number ratio that gives the empirical formula.

Molar mass is divided by the empirical formula mass to know how many empirical formula units are actually present in the compound. The calculations for all the steps are shown below:

calculations for moles:

[tex]283.4gC(\frac{1molC}{12.01gC})[/tex]

= 23.6 mol C

[tex]31.7gH(\frac{1molH}{1.01gH})[/tex]

= 31.4 mol H

[tex]377.4gO(\frac{1molO}{16gO})[/tex]

= 23.6 mol O

Calculations for mol ratio:

In the above moles of C, H and O, the least number is 23.6. So, let's divide each of them by 23.6.

[tex]C=\frac{23.6}{23.6}[/tex] = 1

[tex]H=\frac{31.4}{23.6}[/tex] = 1.33

[tex]O=\frac{23.6}{23.6}[/tex] = 1

1.33 is not a whole number so we need to multiply all of these by 3 that gives almost whole number ratio.

C = 3(1) = 3

H = 3(1.33) = 3.99 = 4

O = 3(1) = 3

So, the empirical formula of the compound is [tex]C_3H_4O_3[/tex] .

Empirical formula mass = 3(12.01)+4(1.01)+3(16)

= 36.03+4.04+48

= 88.07

Number of empirical formula units = [tex]\frac{176.124}{88.07}[/tex] = 2

So, molecular formula of the compound contains two empirical formula units and the molecular formula is [tex]C_6H_8O_6[/tex] .

sarajevo sarajevo · Answer 2 · 2018-10-21T07:57:17+02:00

Mass of carbon = 283.4 grams

Atomic mass of carbon = 12 amu

Moles = mass/ atomic mass

Moles of carbon = 283.4 grams/12 amu

Moles of carbon = 23.62 mol

Mass of hydrogen = 31.7 grams

Atomic mass of hydrogen = 1 amu

Moles of hydrogen = 31.7 grams/1 amu

Moles of hydrogen = 37.1 mol

Mass of oxygen = 377.4 grams

Atomic mass of oxygen = 16 amu

Moles of oxygen = 377.4 grams/16 amu

Moles of oxygen = 23.58 mol

Dividing the number of moles of C, H and O by the least number of moles

C = 23.62 / 23.58 = 1

H = 37.1 / 23.58 = 1.6

O = 23.58 / 23.58 = 1

Multiply the ratio values to find near whole numbers:

C = 1 x 2 = 2

H = 1.6 x 2 = 3.2

O = 1 x 2 = 2

Round the values to nearest whole number:

C = 2

H = 3

O = 2

Thus the empirical formula is C₂H₃O₂

Molar mass of the compound = 176.124 g/mol

Empirical formula mass = (2 x 12 + 1 x 3 + 2 x 16) = 59

n = Molar mass/ Empirical formula mass

n = 176.124 / 59

n = 2.985 = 3

Molecular formula = n x Empirical formula

Molecular formula = 3 x (C₂H₃O₂)

Thus the molecular formula of the compound is C₆H₉O₆