Question #29a4c

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Question #29a4c

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Answer 1 · 2014-09-25T15:37:26

$7.31 \times 10^{- 23}$ $7.31 xx 10^(-23)$ $g$ $"g"$

Explanation:

In order to determine the mass of one molecule of ${CO}_{2}$ $"CO"_2$ , the process involves using mole and mass conversion factors to convert from molecules to grams by way of moles.

molecule $\to$ $->$ mole $\to$ $->$ grams

The conversion factors include:

$6.02 \times 10^{23}$ $6.02 xx 10^23$ $molecules = 1 mole$ $"molecules = 1 mole" " "$ (this is Avogadro's constant)

and

$1 mole = the molar mass of the molecule$ $"1 mole = the molar mass of the molecule"$

The molar mass of carbon dioxide, ${CO}_{2}$ $"CO"_2$ , can be determined by combining the mass of one carbon atom and 2 oxygen atoms.

$C = 1 \times 12.01 g/mol = 12.01 g/mol$ $C = 1 xx "12.01 g/mol" = "12.01 g/mol"$

$O = 2 \times 16.0 g/mol = 32.0 g/mol$ $O = 2 xx "16.0 g/mol" = "32.0 g/mol"$

So

$12.01 g/mol + 32.0 g/mol = 44.01 g/mol$ $"12.01 g/mol + 32.0 g/mol = 44.01 g/mol"$

The conversion is completed as follows:

$1 cancel("molecule CO"_2) xx (1 cancel("mole CO"_2)) / (6.02 xx 10^23 cancel("molecules")) xx "44.01 g"/(1 cancel("mole CO"_2))$

$= 7.31 xx 10^(-23)$ $"g CO"_2$

Molecules cancel with molecules and moles cancel with moles
the outcome is the mass in grams.

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Question #29a4c

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