Question

what is the percent composition of elements in caffeine c₈h₁₀n₄o₂?
percent composition section
10pt
if the percent composition of a compound is 5.75% h, 22.83% o, 19.99% n, and 51.42% c, what is the empirical formula?
if the true molecular weight of that compound is 140.14g/mol, what is the true molecular formula?
balance these equations
balancing 8pt
na₃po₄ + koh → naoh + k₃po₄
c₈h₁₈ + o₂ → co₂ + h₂o
na₃po₄ + cacl₂ → nacl + ca₃(po₄)₂

Explanation:

First Set: Percent Composition & Formulas

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Part 1: Percent Composition of Caffeine ($\text{C}_8\text{H}_{10}\text{N}_4\text{O}_2$)

Step1: Calculate molar mass of caffeine

First, find atomic masses: $\text{C}=12.01$, $\text{H}=1.008$, $\text{N}=14.01$, $\text{O}=16.00$.

$$\begin{align*} \text{Molar mass} &= (8 \times 12.01) + (10 \times 1.008) + (4 \times 14.01) + (2 \times 16.00) \\ &= 96.08 + 10.08 + 56.04 + 32.00 \\ &= 194.20\ \text{g/mol} \end{align*}$$

Step2: % of Carbon

$$\% \text{C} = \frac{8 \times 12.01}{194.20} \times 100 = \frac{96.08}{194.20} \times 100 \approx 49.47\%$$

Step3: % of Hydrogen

$$\% \text{H} = \frac{10 \times 1.008}{194.20} \times 100 = \frac{10.08}{194.20} \times 100 \approx 5.19\%$$

Step4: % of Nitrogen

$$\% \text{N} = \frac{4 \times 14.01}{194.20} \times 100 = \frac{56.04}{194.20} \times 100 \approx 28.86\%$$

Step5: % of Oxygen

$$\% \text{O} = \frac{2 \times 16.00}{194.20} \times 100 = \frac{32.00}{194.20} \times 100 \approx 16.48\%$$
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Part 2: Empirical Formula from % Composition

Step1: Convert % to moles (assume 100g sample)

$$\begin{align*} \text{Moles of C} &= \frac{51.42}{12.01} \approx 4.28\ \text{mol} \\ \text{Moles of H} &= \frac{5.75}{1.008} \approx 5.70\ \text{mol} \\ \text{Moles of N} &= \frac{19.99}{14.01} \approx 1.43\ \text{mol} \\ \text{Moles of O} &= \frac{22.83}{16.00} \approx 1.43\ \text{mol} \end{align*}$$

Step2: Divide by smallest mole value

Smallest mole = 1.43 mol

$$\begin{align*} \text{C}: \frac{4.28}{1.43} &\approx 3 \\ \text{H}: \frac{5.70}{1.43} &\approx 4 \\ \text{N}: \frac{1.43}{1.43} &= 1 \\ \text{O}: \frac{1.43}{1.43} &= 1 \end{align*}$$

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Part 3: Molecular Formula from Empirical Formula

Step1: Calculate empirical formula mass

Empirical formula = $\text{C}_3\text{H}_4\text{NO}$

$$ \text{Empirical mass} = (3 \times 12.01) + (4 \times 1.008) + 14.01 + 16.00 = 70.08\ \text{g/mol} $$

Step2: Find ratio of molecular to empirical mass

$$n = \frac{\text{Molecular mass}}{\text{Empirical mass}} = \frac{140.14}{70.08} \approx 2$$

Step3: Multiply empirical formula by n

$$\text{Molecular formula} = 2 \times (\text{C}_3\text{H}_4\text{NO})$$
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Second Set: Balancing Chemical Equations

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Equation 1: $\text{Na}_3\text{PO}_4 + \text{KOH}

ightarrow \text{NaOH} + \text{K}_3\text{PO}_4$

Step1: Balance K atoms

Add coefficient 3 to KOH:

$$\text{Na}_3\text{PO}_4 + 3\text{KOH} ightarrow \text{NaOH} + \text{K}_3\text{PO}_4$$

Step2: Balance Na and OH atoms

Add coefficient 3 to NaOH:

$$\text{Na}_3\text{PO}_4 + 3\text{KOH} ightarrow 3\text{NaOH} + \text{K}_3\text{PO}_4$$

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Equation 2: $\text{C}_8\text{H}_{18} + \text{O}_2

ightarrow \text{CO}_2 + \text{H}_2\text{O}$

Step1: Balance C and H atoms

Add coefficient 8 to $\text{CO}_2$, 9 to $\text{H}_2\text{O}$:

$$\text{C}_8\text{H}_{18} + \text{O}_2 ightarrow 8\text{CO}_2 + 9\text{H}_2\text{O}$$

Step2: Balance O atoms

Total O on right = $(8 \times 2) + 9 = 25$, so add $\frac{25}{2}$ to $\text{O}_2$, then multiply all coefficients by 2 to eliminate fractions:

$$2\text{C}_8\text{H}_{18} + 25\text{O}_2 ightarrow 16\text{CO}_2 + 18\text{H}_2\text{O}$$

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Equation 3: $\text{Na}_3\text{PO}_4 + \text{CaCl}_2

ightarrow \text{NaCl} + \text{Ca}_3(\text{PO}_4)_2$

Step1: Balance $\text{PO}_4$ and Ca atoms

Add coefficient 2 to $\text{Na}_3\text{PO}_4$, 3 to $\text{CaCl}_2$:

$$2\text{Na}_3\text{PO}_4 + 3\text{CaCl}_2 ightarrow \text{NaCl} + \text{Ca}_3(\text{PO}_4)_2$$

Step2: Balance Na and Cl atoms

Add coefficient 6 to NaCl:

$$2\text{Na}_3\text{PO}_4 + 3\text{CaCl}_2 ightarrow 6\text{NaCl} + \text{Ca}_3(\text{PO}_4)_2$$

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Answer:

Percent Composition & Formulas

1. Percent composition of caffeine:

• Carbon: 49.47%
• Hydrogen: 5.19%
• Nitrogen: 28.86%
• Oxygen: 16.48%

1. Empirical formula: $\text{C}_3\text{H}_4\text{NO}$
2. Molecular formula: $\text{C}_6\text{H}_8\text{N}_2\text{O}_2$

Balanced Chemical Equations

1. $\boldsymbol{1}\text{Na}_3\text{PO}_4 + \boldsymbol{3}\text{KOH}

ightarrow \boldsymbol{3}\text{NaOH} + \boldsymbol{1}\text{K}_3\text{PO}_4$

1. $\boldsymbol{2}\text{C}_8\text{H}_{18} + \boldsymbol{25}\text{O}_2

ightarrow \boldsymbol{16}\text{CO}_2 + \boldsymbol{18}\text{H}_2\text{O}$

1. $\boldsymbol{2}\text{Na}_3\text{PO}_4 + \boldsymbol{3}\text{CaCl}_2

ightarrow \boldsymbol{6}\text{NaCl} + \boldsymbol{1}\text{Ca}_3(\text{PO}_4)_2$