Chemistry, asked by parveenstarP6386, 1 year ago

A compound contains 32% carbon, 4% hydrogen and rest oxygen ,its vapour density is 75. calculate the empirical and molecular formula.

Answers

Answered by mohitahlawat
3
CH2O....... Is both empirical and molecular formula
Answered by kobenhavn
24

Answer: The empirical formula is C_2H_3O_3   and molecular formula is  C_4H_6O_6.

Solution : Given,

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 = 32 g

Mass of H = 4 g

Mass of O = (100-36)= 64 g

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{32g}{12g/mole}=2.66moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4g}{1g/mole}=4moles[/tex]

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{64g}{16g/mole}=4mole

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

For C = \frac{2.66}{2.66}=1

For H = \frac{4}{2.66}=1.5

For O= \frac{4}{2.66}=1.5

The ratio of C : H : O= 1 :1.5 :1.5

Converting them into simple whole number ratio by multiplying by 2.

Hence the empirical formula is C_2H_3O_3  

The empirical weight of C_2H_3O_3  = 2(12) + 3(1) +3(16)= 75g.

Molecular mass=2\times vapor density=2\times 75=150g

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}

n=\frac{150g/mole}{75g/eq}=2

Molecular formula is C_4H_6O_6

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