Write the molecular formula for a compound with the possible elements C, H, N and O that exhibits a molecular ion at M+ = 84.095

1 answer:

8 0

Use the rule of 13 - the formula of a compound is a multiple n of 13 (the molar mass of CH) plus a remainder r. For O, add O and remove CH₄ and for N, add N and remove one CH₂.
84 ÷ 13 = 6r6, the integer equals the number of carbon atoms and the integer plus the remainder equals the number of hydrogen atoms, so possible formula is C₆H₁₂.
M(C₆H₁₂) = 6 · 12,00 + 12 · 1,008 = 84,096.
Add one oxygen atom to molecular formula: C₅H₈O.
M(C₅H₈O) = 5 · 12,00 + 8 · 1,008 + 15,995 = 84,059, not.
Add one nitrogen atom to molecular formula: C₅H₁₀N.
M(C₅H₁₀N) = 5 · 12,00 + 10 · 1,008 + 14,003 = 84,083, not.
If we add nitrogen and oxygen mass will be lower than 84,095, so molecular formula is C₆H₁₂.

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CH4 + 202 → CO2 + 2H2O How many grams of O2 needed to produce 36 grams of H2O?

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<h3>Answer:</h3>

64 g O₂

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

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 36 g H₂O

[Solve] x g O₂

Step 2: Identify Conversions

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Stoichiometry

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$