Question

a compund is found to contain 64.80% carbon 13.62% hydrogen and 21.58% oxygen by weight .what is the empirical formula for this compound ?the molecular weight for this compound id 74.14g/mol. what is its molecular formula.
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Answer 1

Answer:

Molecular formula = $C_{4}H_{10}O_{1}$

Explanation:

1) Assume 100 g of the compound is present. This changes the percents to grams:

C ⇒ 64.80 g

H ⇒ 13.62 g

O ⇒ 21.58 g

2) Convert the masses to moles:

C ⇒ 64.80 g / 12 g = 5.4

H ⇒ 13.62 g / 1 g = 13.62

O ⇒ 21.58 g / 16 g = 1.349

3) Divide by the lowest, for getting the smallest whole-number ratio:

C ⇒ 5.4 / 1.349 = 4

H ⇒ 13.62 / 1.349 = 10

O ⇒ 1.349 / 1.349 = 1

4) Write the empirical formula:

$C_{4}H_{10}O_{1}$

5) Determine the molecular formula:

Now empirical formula`s mass:-

=12×4 + 1×10 + 16×1

=48+10+16

=74

Now we know that;

molecular weight = n × empirical weight

∴ 74.14 = n × 74

∴ n ≅ 1.00

Now we also know that

molecular formula = n × empirical formula

∴ molecular formula = 1 × $C_{4}H_{10}O_{1}$

∴ Molecular formula:- $C_{4}H_{10}O_{1}$

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