Chemistry Examples

$Z n {(C H_{3} C O O)}_{2}$

Step 1

To find the mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2}$ 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 $Z n {(C H_{3} C O O)}_{2}$ $=$ $(mass of Zn)$ + $2 (mass of C)$ + $6 (mass of H)$ + $2 (mass of C)$ + $2 (mass of O)$ + $2 (mass of O)$

Step 2

Fill in the atomic masses from the periodic table.

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2}$ $=$ $(65.38999938)$ + $2 (12.01000022)$ + $6 (1.00999999)$ + $2 (12.01000022)$ + $2 (16)$ + $2 (16)$

Step 3

Simplify the result.

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Step 3.1

Simplify each term.

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Step 3.1.1

Multiply $2$ by $12.01000022$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 65.38999938 + 24.02000045 + 6 (1.00999999) + 2 (12.01000022) + 2 (16) + 2 (16) g$

Step 3.1.2

Multiply $6$ by $1.00999999$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 65.38999938 + 24.02000045 + 6.05999994 + 2 (12.01000022) + 2 (16) + 2 (16) g$

Step 3.1.3

Multiply $2$ by $12.01000022$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 65.38999938 + 24.02000045 + 6.05999994 + 24.02000045 + 2 (16) + 2 (16) g$

Step 3.1.4

Multiply $2$ by $16$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 65.38999938 + 24.02000045 + 6.05999994 + 24.02000045 + 32 + 2 (16) g$

Step 3.1.5

Multiply $2$ by $16$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 65.38999938 + 24.02000045 + 6.05999994 + 24.02000045 + 32 + 32 g$

Step 3.2

Simplify by adding numbers.

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Step 3.2.1

Add $65.38999938$ and $24.02000045$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 89.40999984 + 6.05999994 + 24.02000045 + 32 + 32 g$

Step 3.2.2

Add $89.40999984$ and $6.05999994$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 95.46999979 + 24.02000045 + 32 + 32 g$

Step 3.2.3

Add $95.46999979$ and $24.02000045$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 119.49000024 + 32 + 32 g$

Step 3.2.4

Add $119.49000024$ and $32$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 151.49000024 + 32 g$

Step 3.2.5

Add $151.49000024$ and $32$ .

mass of $1$ mole of $Z n {(C H_{3} C O O)}_{2} = 183.49000024 g$