# How many moles are present in a "351 kg" block of calcium?

Sep 21, 2016

The number of moles in $\text{351 kg Ca}$ is $8.76 \times {10}^{3}$.

#### Explanation:

First determine the mass of one mole (i.e. molar mass) of calcium. The molar mass of an element is its atomic weight on the periodic table in g/mol.

The molar mass of calcium is 40.078 g/mol .

Next, since molar mass is g/mol, we need to convert 351 kg to g. There are 1000 g in 1 kg.

$351 \cancel{\text{kg Ca"xx(1000 "g Ca")/(1cancel"kg Ca")="351 000 g}}$

$\text{351 kg Ca=351 000 g Ca}$

Next we'll convert the given mass to moles by dividing the given mass by the molar mass.

$351 000 \cancel{\text{g Ca"xx(1"mol Ca")/(40.078cancel"g Ca")=8.76xx10^3 "mol Ca}}$ (rounded to three significant figures)

Sep 23, 2016

There are approx. $9000$ moles of calcium metal.

#### Explanation:

$\text{Calcium metal, } Z = 20 ,$ has a molar mass of $40.08 \cdot g \cdot m o {l}^{-} 1$. From where did I get this figure? From where will get it if you are asked this on an exam? You will not be expected to remember it.

And molar quantity is simply the quotient:

$=$ $\text{Mass"/"Molar mass}$ $=$ $\frac{351.0 \times {10}^{3} \cdot \cancel{g}}{40.08 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ ??mol

Note that this calculation is dimensionally consistent. I wanted an answer in $\text{moles}$. The mass units cancelled out so I got an answer in $\frac{1}{m o {l}^{-} 1} = \text{moles}$. This persuades me that I set up the question right (for once!). Are you happy with this treatment?