Question

A compound with a molar mass of 544.0 g/mol is made up of 26.5 grams of Carbon, 2.94 grams
Hydrogen, and 70.6 grams of Oxygen. What is its empirical and molecular formula?

Answer 1

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Answer 2

Given,

the molar mass of the given compound = 544 g/mol

mass of C = 26.5 g

mass of H = 2.94 g

mass of O = 70.6 g

To find,

Empirical formula and the molecular formula of the compound.

Solution,

Dividing the mass elements by their molar mass

C = 26.5 / 12 = 2.208

H = 2.94 / 1 = 2.94

O = 70.6 / 16 = 4.4125

Dividing all by the smallest

C = 2.208 / 2.208 = 1

H = 2.94 / 2.208 = 1.33

O = 4.4125 / 2.208 = 2

Muitiplying to express in whole numbers (by 3)

C = 1 × 3 = 3

H = 1.33 × 3 = 4

O = 2 × 3 = 6

∴Empirical formula = $C_3H_4O_6$

∵Empirical formula = $C_3H_4O_6$

Molecular formula = $[C_3H_4O_6]_n$

$[C_3H_4O_6]_n$ = 544

⇒ [(12×3) + (4×1) + (16×6)]n = 544

⇒ [36 + 4 + 96]n = 544

⇒ 136n = 544

⇒n = 544 / 136

⇒n = 4

⇒Molecular formula = $[C_3H_4O_6]_4$

∴Molecular formula = C₁₂H₁₆O₂₄