Question

1. kmno₄ 1×39.1 + 1×54.9 + 4×16.0 = 39.1 + 54.9 + 64.0 = 158.0 g/mol

% k = (39.1 / 158) × 100 = 24.7 %
% mn = (54.9 / 158) × 100 = 34.7 %
% o = (64.0 / 158) × 100 = 40.5 %
k: 24.7%
mn: 34.7%
o: 40.5%

1. hcl

% h = ____ × 100 =
% cl = ____ × 100 =
h: ____
cl: ____

1. nano₃

% na = ____ × 100 =
% n = ____
% o = ____
na: ____
n: ____
o: ____

1. nh₄co₃

% n =
% h =
% o =
% c =
n: ____
h: ____
o: ____
c: ____

Explanation:

Problem 6: HCl

Step 1: Calculate molar mass of HCl

Molar mass of H = 1.0 g/mol, molar mass of Cl = 35.5 g/mol.
Molar mass of HCl = \( 1.0 + 35.5 = 36.5 \, \text{g/mol} \)

Step 2: Calculate %H

Mass of H in HCl = 1.0 g/mol.
\( \% \text{H} = \frac{1.0}{36.5} \times 100 \approx 2.74\% \)

Step 3: Calculate %Cl

Mass of Cl in HCl = 35.5 g/mol.
\( \% \text{Cl} = \frac{35.5}{36.5} \times 100 \approx 97.26\% \)

Step 1: Calculate molar mass of NaNO₃

Molar mass of Na = 23.0 g/mol, N = 14.0 g/mol, O = 16.0 g/mol.
Molar mass of NaNO₃ = \( 23.0 + 14.0 + (3 \times 16.0) = 23.0 + 14.0 + 48.0 = 85.0 \, \text{g/mol} \)

Step 2: Calculate %Na

Mass of Na in NaNO₃ = 23.0 g/mol.
\( \% \text{Na} = \frac{23.0}{85.0} \times 100 \approx 27.06\% \)

Step 3: Calculate %N

Mass of N in NaNO₃ = 14.0 g/mol.
\( \% \text{N} = \frac{14.0}{85.0} \times 100 \approx 16.47\% \)

Step 4: Calculate %O

Mass of O in NaNO₃ = \( 3 \times 16.0 = 48.0 \, \text{g/mol} \).
\( \% \text{O} = \frac{48.0}{85.0} \times 100 \approx 56.47\% \)

Step 1: Molar mass of \( (\text{NH}_4)_2\text{CO}_3 \)

Molar mass: N = 14.0, H = 1.0, C = 12.0, O = 16.0.
\( (\text{NH}_4)_2\text{CO}_3 = 2(14.0 + 4 \times 1.0) + 12.0 + 3 \times 16.0 = 2(18.0) + 12.0 + 48.0 = 36.0 + 12.0 + 48.0 = 96.0 \, \text{g/mol} \)

Step 2: Calculate %N

Mass of N: \( 2 \times 14.0 = 28.0 \, \text{g/mol} \).
\( \% \text{N} = \frac{28.0}{96.0} \times 100 \approx 29.17\% \)

Step 3: Calculate %H

Mass of H: \( 2 \times 4 \times 1.0 = 8.0 \, \text{g/mol} \).
\( \% \text{H} = \frac{8.0}{96.0} \times 100 \approx 8.33\% \)

Step 4: Calculate %O

Mass of O: \( 3 \times 16.0 = 48.0 \, \text{g/mol} \).
\( \% \text{O} = \frac{48.0}{96.0} \times 100 = 50.0\% \)

Step 5: Calculate %C

Mass of C: 12.0 g/mol.
\( \% \text{C} = \frac{12.0}{96.0} \times 100 = 12.5\% \)

Answer:

H: \( \boldsymbol{2.74\%} \)
Cl: \( \boldsymbol{97.26\%} \)

Problem 7: NaNO₃