Question

1. determine the molecular formula of a compound that is 49.48% carbon, 5.19% hydrogen, 28.85% nitrogen, and 16.48% oxygen. the molecular weight is 194.19 g/mol. a) c₈h₁₂n₄o₂ b) c₄h₅n₂o c) c₈h₁₀n₄o₂ d) c₈h₁₀n₂o 11) combustion analysis of 63.8 mg of a c, h and o containing compound produced 145.0 mg of co₂ and 59.38 mg of h₂o. what is the empirical formula for the compound? a) c₅h₂o b) cho c) c₃h₆o d) c₃h₇o e) c₆ho₃ 12) radium phosphate reacts with sulfuric acid to form radium sulfate and phosphoric acid. what is the coefficient for sulfuric acid when the equation is balanced using the lowest, whole - numbered coefficients? a) 1 b) 2 c) 3 d) none of these 13) balance the following equation. ____c₁₀h₁₂ + __o₂ → __h₂o + __co₂ 14) balance the following equation. __c₉h₂₀ + __o₂ → __h₂o + ____co₂

Explanation:

Question 10

Step1: Assume 100g of compound

Moles of C: $\frac{49.48\,\text{g}}{12.01\,\text{g/mol}} \approx 4.12\,\text{mol}$
Moles of H: $\frac{5.19\,\text{g}}{1.008\,\text{g/mol}} \approx 5.15\,\text{mol}$
Moles of N: $\frac{28.85\,\text{g}}{14.01\,\text{g/mol}} \approx 2.06\,\text{mol}$
Moles of O: $\frac{16.48\,\text{g}}{16.00\,\text{g/mol}} \approx 1.03\,\text{mol}$

Step2: Divide by smallest mole (O: 1.03)

C: $\frac{4.12}{1.03} \approx 4$
H: $\frac{5.15}{1.03} \approx 5$
N: $\frac{2.06}{1.03} = 2$
O: $\frac{1.03}{1.03} = 1$
Empirical formula: $C_4H_5N_2O$ (molar mass: $4(12.01)+5(1.008)+2(14.01)+16.00 = 97.11\,\text{g/mol}$)

Step3: Find multiplier

$\frac{194.19}{97.11} \approx 2$
Molecular formula: $2 \times (C_4H_5N_2O) = C_8H_{10}N_4O_2$ (Check option C)

Step1: Moles of C from $CO_2$

$n_{CO_2} = \frac{145.0\,\text{mg}}{44.01\,\text{mg/mmol}} \approx 3.295\,\text{mmol}$ → $n_C = 3.295\,\text{mmol}$

Step2: Moles of H from $H_2O$

$n_{H_2O} = \frac{59.38\,\text{mg}}{18.02\,\text{mg/mmol}} \approx 3.295\,\text{mmol}$ → $n_H = 2 \times 3.295 = 6.59\,\text{mmol}$

Step3: Mass of O

Mass of C: $3.295\,\text{mmol} \times 12.01\,\text{mg/mmol} \approx 39.57\,\text{mg}$
Mass of H: $6.59\,\text{mmol} \times 1.008\,\text{mg/mmol} \approx 6.64\,\text{mg}$
Mass of O: $63.8 - 39.57 - 6.64 = 17.59\,\text{mg}$
Moles of O: $\frac{17.59\,\text{mg}}{16.00\,\text{mg/mmol}} \approx 1.099\,\text{mmol}$

Step4: Find mole ratios (divide by O moles: ~1.1)

C: $\frac{3.295}{1.099} \approx 3$
H: $\frac{6.59}{1.099} \approx 6$
O: $\frac{1.099}{1.099} = 1$
Empirical formula: $C_3H_6O$ (Option C)

Step1: Write unbalanced equation

$Ra_3(PO_4)_2 + H_2SO_4
ightarrow RaSO_4 + H_3PO_4$

Step2: Balance Ra (3 on left, 1 on right)

$Ra_3(PO_4)_2 + H_2SO_4
ightarrow 3RaSO_4 + H_3PO_4$

Step3: Balance PO₄ (2 on left, 1 on right)

$Ra_3(PO_4)_2 + H_2SO_4
ightarrow 3RaSO_4 + 2H_3PO_4$

Step4: Balance SO₄ (3 on right, 1 on left)

$Ra_3(PO_4)_2 + 3H_2SO_4
ightarrow 3RaSO_4 + 2H_3PO_4$

Step5: Check H and O (balanced). Coefficient for $H_2SO_4$ is 3 (Option C)

Answer:

C. $C_8H_{10}N_4O_2$

Question 11