Question

A carbon compound contains 12.8% Carbon, 2.1% Hydrogen, 85.1% Bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula.

Answer 1

so, now we know the empirical formula

now,
we know
molecular formula=n×empirical formula
=2× CH2Br
= C2H4Br2 Answer

Answer 2

The molecular formula is  $C_2H_4Br_2$

Explanation:

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

a) If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 12.8 g

Mass of H= 2.1 g

Mass of Br= 85.1 g

Step 1 : convert given masses into moles

Moles of C=$\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{12.8g}{12g/mole}=1.07moles$

Moles of H=$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{2.1g}{1g/mole}=2.1moles$

Moles of Br=$\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{85.1g}{79.9g/mole}=1.07moles$

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

For C = $\frac{1.07}{1.07}=1$

For H =$\frac{2.1}{1.07}=2$

For Br =$\frac{1.07}{1.07}=1$

The ratio of C: H : Br = 1: 2: 1

Hence the empirical formula is $CH_2Br$

Empirical mass =$1\times 12+2\times 1+1\times 79.9=93.9g$

Molecular mass = 187.9

n= $\frac{\text {Molecular mass}{\text {Empirical mass}}=\frac{187.9}{93.9}=2$

Thus molecular formula = $2\times CH_2Br=C_2H_4Br_2$

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