# Question 6c787

Jul 9, 2015

Here's how you can approach such problems.

#### Explanation:

The question asks you determine what the mass of a large group of iron atoms, $150 \cdot {10}^{12}$ to be precise, would be is an individual iron atom has a mass of 56 u.

Now, they tell you that each of those atoms has a mass of 56 u. The symbol "u" is used to denote the equivalent of $\frac{1}{12} ^ \text{th}$ of the mass of a ${\text{_}}^{12} C$ atom.

In more practical terms,

$\text{1 u" = 1.66 * 10^(-24)"g}$

Since each atom of iron weighs 56 times more than that, you have

56cancel("u") * (1.66 * 10^(-24)"g")/(1cancel("u")) = 9.3 * 10^(-23)"g"#

Each iron atom has a mass of $9.3 \cdot {10}^{- 23} \text{g}$. Now all you have to do is multiply the number of atoms with the mass of an individual atom to get

$150 \cdot {10}^{12} \cancel{\text{atoms") * (9.3 * 10^(-23)"g")/(1cancel("atom")) = color(green)(1.4 * 10^(-14)"g}}$

An interesting thing to notice here is that if you use the number of atoms present in 1 mole of iron, and multiply this number by the mass of an individual iron atom, you'll get the molar mass of iron.

By definition, one mole of any substance contains exactly $6.022 \cdot {10}^{23}$ atoms or molecules of that substance - this is known as Avogadro's number.

In your case, one mole of iron would have a mass of

$6.022 \cdot {10}^{23} \text{atoms" * (9.3 * 10^(-24)"g")/(1cancel("atom")) = "56.0 g}$

If you take into account rounding and sig figs, this result matches the actual molar mass of iron, which is 55.845 g.

So remember, one "u" is always equal to $1.66 8 {10}^{- 24} \text{g}$. If you know how the mass of a single atom, you can always determine the mass of a whole bunch of atoms by using that mass.