Chemistry Examples

K O_{2}

$K O_{2}$

Step 1

To find the mass of

1

$1$ mole of

K O_{2}

$K O_{2}$ 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

$1$ mole of

K O_{2}

$K O_{2}$

=

$=$

(mass of K)

$(mass of K)$ +

2 (mass of O)

$2 (mass of O)$

Step 2

Fill in the atomic masses from the periodic table.

mass of

1

$1$ mole of

K O_{2}

$K O_{2}$

=

$=$

(39.09999847)

$(39.09999847)$ +

2 (16)

$2 (16)$

Step 3

Simplify the result.

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Step 3.1

Multiply

2

$2$ by

16

$16$ .

mass of

1

$1$ mole of

K O_{2} = 39.09999847 + 32 g

$K O_{2} = 39.09999847 + 32 g$

Step 3.2

Add

39.09999847

$39.09999847$ and

32

$32$ .

mass of

1

$1$ mole of

K O_{2} = 71.09999847 g

$K O_{2} = 71.09999847 g$

mass of

1

$1$ mole of

K O_{2} = 71.09999847 g

$K O_{2} = 71.09999847 g$

Step 4

One mole of any gas occupies

22.4 L

$22.4 L$ at STP (standard temperature and pressure).

V = 22.4 L

$V = 22.4 L$

Step 5

Density is mass over volume.

\frac{71.09999847 g}{22.4 L}

$\frac{71.09999847 g}{22.4 L}$

Step 6

Factor

71.09999847

$71.09999847$ out of

71.09999847 g

$71.09999847 g$ .

\frac{71.09999847 (g)}{22.4 L}

$\frac{71.09999847 (g)}{22.4 L}$

Step 7

Factor

22.4

$22.4$ out of

22.4 L

$22.4 L$ .

\frac{71.09999847 (g)}{22.4 (L)}

$\frac{71.09999847 (g)}{22.4 (L)}$

Step 8

Separate fractions.

\frac{71.09999847}{22.4} \cdot \frac{g}{L}

$\frac{71.09999847}{22.4} \cdot \frac{g}{L}$

Step 9

Divide

71.09999847

$71.09999847$ by

22.4

$22.4$ .

3.17410707 \frac{g}{L}

$3.17410707 \frac{g}{L}$

Step 10

Combine

3.17410707

$3.17410707$ and

\frac{g}{L}

$\frac{g}{L}$ .

\frac{3.17410707 g}{L}

$\frac{3.17410707 g}{L}$

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