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LuckyWell [14K]
3 years ago
9

What is the empirical formula for a compound that contains 38.77% Cl and 61.23% O?

Chemistry
1 answer:
Leno4ka [110]3 years ago
3 0
W(Cl)=0.3877
w(O)=0.6123
M(Cl)=35.5 g/mol
M(O)=16.0 g/mol

ClₐOₓ
M(ClₐOₓ)=35.5a+16.0x

w(Cl)=35.5a/(35.5a+16.0x)
w(O)=16.0x/(35.5a+16.0x)

solve a system of two equations with two unknowns
35.5a/(35.5a+16.0x)=0.3877
16.0x/(35.5a+16.0x)=0.6123

a=2
x=7

Cl₂O₇ is the empirical formula
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3 years ago
A compound has a percent composition of 38.7% carbon, 9.76% hydrogen, 51.5% oxygen.
goldenfox [79]

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)

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  • 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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The molarity (M) of an aqueous solution containing 29.8 g of sucrose, C12H22O11, in 120 mL of solution is:
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Answer:

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Step-by-step explanation:

c = moles/litres

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Litres = 120 × 1/1000

Litres = 0.120 L

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c = 0.725 mol·L⁻¹


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