Chemistry Examples

$C_{2} H_{20} N C l_{3} B r_{5}$

Step 1

To find the mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5}$ look up the atomic mass of each element and multiply it by the number of atoms contained in each element in the molecule.

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5}$ $=$ $2 (mass of C)$ + $20 (mass of H)$ + $(mass of N)$ + $3 (mass of Cl)$ + $5 (mass of Br)$

Step 2

Fill in the atomic masses from the periodic table.

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5}$ $=$ $2 (12.01000022)$ + $20 (1.00999999)$ + $(14.01000022)$ + $3 (35.45000076)$ + $5 (79.90000152)$

Step 3

Simplify the result.

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Step 3.1

Remove parentheses.

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5} = 2 (12.01000022) + 20 (1.00999999) + 14.01000022 + 3 (35.45000076) + 5 (79.90000152) g$

Step 3.2

Simplify each term.

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Step 3.2.1

Multiply $2$ by $12.01000022$ .

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5} = 24.02000045 + 20 (1.00999999) + 14.01000022 + 3 (35.45000076) + 5 (79.90000152) g$

Step 3.2.2

Multiply $20$ by $1.00999999$ .

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5} = 24.02000045 + 20.1999998 + 14.01000022 + 3 (35.45000076) + 5 (79.90000152) g$

Step 3.2.3

Multiply $3$ by $35.45000076$ .

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5} = 24.02000045 + 20.1999998 + 14.01000022 + 106.35000228 + 5 (79.90000152) g$

Step 3.2.4

Multiply $5$ by $79.90000152$ .

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5} = 24.02000045 + 20.1999998 + 14.01000022 + 106.35000228 + 399.50000762 g$

Step 3.3

Simplify by adding numbers.

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Step 3.3.1

Add $24.02000045$ and $20.1999998$ .

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5} = 44.22000026 + 14.01000022 + 106.35000228 + 399.50000762 g$

Step 3.3.2

Add $44.22000026$ and $14.01000022$ .

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5} = 58.23000049 + 106.35000228 + 399.50000762 g$

Step 3.3.3

Add $58.23000049$ and $106.35000228$ .

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5} = 164.58000278 + 399.50000762 g$

Step 3.3.4

Add $164.58000278$ and $399.50000762$ .

mass of $1$ mole of $C_{2} H_{20} N C l_{3} B r_{5} = 564.08001041 g$

Step 4

Find the mass percentage of $C$ .

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Step 4.1

The mass composition percentage is the mass of $C$ divided by the mass of the entire molecule converted to a percentage.

Mass percentage of $C = \frac{24.02}{564.08001041} \cdot 100$

Step 4.2

Divide $24.02$ by $564.08001041$ .

$0.04258261 \cdot 100$

Step 4.3

Multiply $0.04258261$ by $100$ .

$4.25826116$

Step 4.4

Set the number as a percent.

Mass percentage of $C = 4.25826116 %$

Step 5

Find the mass percentage of $H$ .

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Step 5.1

The mass composition percentage is the mass of $H$ divided by the mass of the entire molecule converted to a percentage.

Mass percentage of $H = \frac{20.2}{564.08001041} \cdot 100$

Step 5.2

Divide $20.2$ by $564.08001041$ .

$0.03581052 \cdot 100$

Step 5.3

Multiply $0.03581052$ by $100$ .

$3.58105226$

Step 5.4

Set the number as a percent.

Mass percentage of $H = 3.58105226 %$

Step 6

Find the mass percentage of $N$ .

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Step 6.1

The mass composition percentage is the mass of $N$ divided by the mass of the entire molecule converted to a percentage.

Mass percentage of $N = \frac{14.01}{564.08001041} \cdot 100$

Step 6.2

Divide $14.01$ by $564.08001041$ .

$0.0248369 \cdot 100$

Step 6.3

Multiply $0.0248369$ by $100$ .

$2.48369021$

Step 6.4

Set the number as a percent.

Mass percentage of $N = 2.48369021 %$

Step 7

Find the mass percentage of $C l$ .

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Step 7.1

The mass composition percentage is the mass of $C l$ divided by the mass of the entire molecule converted to a percentage.

Mass percentage of $C l = \frac{106.35}{564.08001041} \cdot 100$

Step 7.2

Divide $106.35$ by $564.08001041$ .

$0.18853708 \cdot 100$

Step 7.3

Multiply $0.18853708$ by $100$ .

$18.85370834$

Step 7.4

Set the number as a percent.

Mass percentage of $C l = 18.85370834 %$

Step 8

Find the mass percentage of $B r$ .

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Step 8.1

The mass composition percentage is the mass of $B r$ divided by the mass of the entire molecule converted to a percentage.

Mass percentage of $B r = \frac{399.5}{564.08001041} \cdot 100$

Step 8.2

Divide $399.5$ by $564.08001041$ .

$0.70823286 \cdot 100$

Step 8.3

Multiply $0.70823286$ by $100$ .

$70.82328616$

Step 8.4

Set the number as a percent.

Mass percentage of $B r = 70.82328616 %$

Step 9

Find the mass percentage of each element in the molecule.

Mass percentage of $C = 4.25826116 %$

Mass percentage of $H = 3.58105226 %$

Mass percentage of $N = 2.48369021 %$

Mass percentage of $C l = 18.85370834 %$

Mass percentage of $B r = 70.82328616 %$