What is the mass of 1.2*10^181.2⋅1018 formula units of calcium chloride?

1 Answer
Stefan V.
Mar 20, 2018

2.2 * 10^(-4)2.2⋅10−4 "g"g

Explanation:

The thing to keep in mind about formula units is that you need 6.022 * 10^(23)6.022⋅1023 of them to have exactly 11 mole of an ionic compound ->→ think Avogadro's constant here.

In this case, you know that 6.022 * 10^(23)6.022⋅1023 formula units of calcium chloride are needed in order to have exactly 11 mole of calcium chloride.

Moreover, you know that calcium chloride has a molar mass of "110.98 g mol"^(-1)110.98 g mol−1, which means that 11 mole of calcium chloride has a mass of "110.98 g"110.98 g.

So if 11 mole of calcium chloride contains 6.022 * 10^(23)6.022⋅1023 formula units and has a mass of "110.98 g"110.98 g, you can say that 6.022 * 10^(23)6.022⋅1023 formula units of calcium chloride have a mass of "110.98 g"110.98 g.

This means that your sample has a mass of

1.2 * 10^8 color(red)(cancel(color(black)("f. units CaCl"_2))) * "110.98 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("f. units CaCl"_2)))) = color(darkgreen)(ul(color(black)(2.2 * 10^(-4) quad "g")))

The answer is rounded to two sig figs, the number of sig figs you have for the number of formula units.

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