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Paul [167]
3 years ago
15

A laboratory analysis of an unknown sample yields 40.5% carbon, 4.5% hydrogen, and 55.0% oxygen. if the sample has a molar mass

of 176.0 g/mol, what is the molecular formula of the sample
Chemistry
1 answer:
den301095 [7]3 years ago
6 0
Answer is: molecular formula is C₆H₈O₆.
n(C) = m(C) ÷ M(C).
n(C) = 40,5 g ÷ 12 g/mol.
n(C) = 3,375 mol.
n(H) = m(H) ÷ M(H).
n(H) = 4,5 g ÷ 1 g/mol.
n(H) = 4,5 mol.
n(O) = m(O) ÷ M(O).
n(O) = 55 g ÷ 16 g/mol.
n(O) = 3,4 mol.
n(C) : n(H) : n(O) = 3,375 mol : 4,5 mol : 3,4 mol / :3,375.
n(C) : n(H) : n(O) = 1 : 1,33 : 1.
n(C) : n(H) : n(O) = 3 : 4 : 3.
M(C₃H₄O₃) = 88 g/mol · 2 = 176 g/mol.

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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.0 mol H₂O

<u>Step 2: Identify Conversions</u>

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<u>Step 3: Convert</u>

<u />\displaystyle 50.0 \ mol \ H_2O(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O} ) = 3.011 × 10²⁵ molecules H₂O

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O

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