A compound has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen.

1 answer:

7 0

The molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.

<h3>How to calculate molecular formula?</h3>

The molecular formula can be calculated from the empirical formula. The empirical formula of the compound is calculated as follows:

C = 38.7% = 38.7g
H = 9.76% = 9.76g
O = 51.5% = 51.5g

Next, we convert the mass to moles by dividing by their atomic mass:

C = 38.7 ÷ 12 = 3.23mol
H = 9.76 ÷ 1 = 9.76mol
O = 51.5÷ 16 = 3.22mol

Next, we divide by the smallest (3.22)

C = 1
H = 3
O = 1

Hence, the empirical formula of the compound is CH3O

If the molar mass of the compound is 62g/mol;

(CH3O)n = 62

31n = 62

n = 2

(CH3O)2 = C2H6O2

Therefore, the molecular formula of the compound that has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen is C2H6O2.

Learn more about molecular formula at: brainly.com/question/14425592

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A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was fo

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Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

$\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M$