Question

four: percent composition
objectives: swbat determine the percent composition of a given compound. swbat use dimensional analysis to convert between moles, masses, and particles.
essential question: what is percent composition and how is it related to the chemical make-up of a compound?
____________________________________; the amount (as a percent) of each element in a compound.
potassium dichromate ($k_2cr_2o_7$) is 35.4% cr, 38.1% o and 26.5% k
$\text{percent (\\%) composition} = \frac{\text{part}}{\text{total}} \times 100 = \frac{\text{mass of atom}}{\text{total mass of compound}} \times 100$
example 1: find the percent composition by mass of a compound that contains 2 g of h and 16 g of o.
total mass of the compound = ______ + ____ = ______
% hydrogen (h) = ________ $\times 100$ =
% oxygen (o) = ________ $\times 100$ =
example 2: find the percent composition by mass of $nh_3$
total mass of the compound = molar mass = ________
% n = ________ $\times 100$ =
% h = ________ $\times 100$ =
more practice:

1. a sample of an unknown compound is made of 0.428g of fluorine and 0.419g of iron. what is the percent composition by mass?

total mass of the compound = ______ + ____ = ______
% f =
% fe =

1. determine the percentage composition by mass of $al_2o_3$.

total mass of the compound = molar mass = ________
% al =
% o =

1. what is the percentage composition by mass of $c_2h_6o$?

Explanation:

Example 1

Step1: Calculate total compound mass

Add masses of H and O:
$\text{Total Mass} = 2\ \text{g} + 16\ \text{g} = 18\ \text{g}$

Step2: Find % of Hydrogen

Divide H mass by total mass, multiply by 100:
$\% \text{H} = \frac{2}{18} \times 100 \approx 11.11\%$

Step3: Find % of Oxygen

Divide O mass by total mass, multiply by 100:
$\% \text{O} = \frac{16}{18} \times 100 \approx 88.89\%$

Example 2

Step1: Calculate molar mass of $\text{NH}_3$

Use atomic masses: $\text{N}=14\ \text{g/mol}$, $\text{H}=1\ \text{g/mol}$
$\text{Total Mass} = 14 + (3 \times 1) = 17\ \text{g/mol}$

Step2: Find % of Nitrogen

Divide N mass by total mass, multiply by 100:
$\% \text{N} = \frac{14}{17} \times 100 \approx 82.35\%$

Step3: Find % of Hydrogen

Divide total H mass by total mass, multiply by 100:
$\% \text{H} = \frac{3}{17} \times 100 \approx 17.65\%$

More Practice 1

Step1: Calculate total compound mass

Add masses of F and Fe:
$\text{Total Mass} = 0.428\ \text{g} + 0.419\ \text{g} = 0.847\ \text{g}$

Step2: Find % of Fluorine

Divide F mass by total mass, multiply by 100:
$\% \text{F} = \frac{0.428}{0.847} \times 100 \approx 50.53\%$

Step3: Find % of Iron

Divide Fe mass by total mass, multiply by 100:
$\% \text{Fe} = \frac{0.419}{0.847} \times 100 \approx 49.47\%$

More Practice 2

Step1: Calculate molar mass of $\text{Al}_2\text{O}_3$

Use atomic masses: $\text{Al}=27\ \text{g/mol}$, $\text{O}=16\ \text{g/mol}$
$\text{Total Mass} = (2 \times 27) + (3 \times 16) = 54 + 48 = 102\ \text{g/mol}$

Step2: Find % of Aluminum

Divide total Al mass by total mass, multiply by 100:
$\% \text{Al} = \frac{54}{102} \times 100 \approx 52.94\%$

Step3: Find % of Oxygen

Divide total O mass by total mass, multiply by 100:
$\% \text{O} = \frac{48}{102} \times 100 \approx 47.06\%$

More Practice 3

Step1: Calculate molar mass of $\text{C}_2\text{H}_6\text{O}$

Use atomic masses: $\text{C}=12\ \text{g/mol}$, $\text{H}=1\ \text{g/mol}$, $\text{O}=16\ \text{g/mol}$
$\text{Total Mass} = (2 \times 12) + (6 \times 1) + 16 = 24 + 6 + 16 = 46\ \text{g/mol}$

Step2: Find % of Carbon

Divide total C mass by total mass, multiply by 100:
$\% \text{C} = \frac{24}{46} \times 100 \approx 52.17\%$

Step3: Find % of Hydrogen

Divide total H mass by total mass, multiply by 100:
$\% \text{H} = \frac{6}{46} \times 100 \approx 13.04\%$

Step4: Find % of Oxygen

Divide O mass by total mass, multiply by 100:
$\% \text{O} = \frac{16}{46} \times 100 \approx 34.78\%$

Answer:

Example 1

Total Mass of the compound = $2\ \text{g} + 16\ \text{g} = 18\ \text{g}$
$\% \text{Hydrogen (H)} = \frac{2}{18} \times 100 \approx 11.11\%$
$\% \text{Oxygen (O)} = \frac{16}{18} \times 100 \approx 88.89\%$

Example 2

Total Mass of the compound = molar mass = $17\ \text{g/mol}$
$\% \text{N} = \frac{14}{17} \times 100 \approx 82.35\%$
$\% \text{H} = \frac{3}{17} \times 100 \approx 17.65\%$

More Practice 1

Total Mass of the compound = $0.428\ \text{g} + 0.419\ \text{g} = 0.847\ \text{g}$
$\% \text{F} \approx 50.53\%$
$\% \text{Fe} \approx 49.47\%$

More Practice 2

Total Mass of the compound = molar mass = $102\ \text{g/mol}$
$\% \text{Al} \approx 52.94\%$
$\% \text{O} \approx 47.06\%$

More Practice 3

Total Mass of the compound = molar mass = $46\ \text{g/mol}$
$\% \text{C} \approx 52.17\%$, $\% \text{H} \approx 13.04\%$, $\% \text{O} \approx 34.78\%$