# Question a95a2

Mar 3, 2015

Let's start with the number of nickel atoms, since it's the easiest one to calculate.

Before determining how many atoms of nickel you have in 1 gram, you must first determine how many moles of nickel you have. This is done by using its molar mass, 58.7 g/mol.

$\text{1 g" * "1 mole"/"58.7 g" = "0.0170 moles Ni}$

Use Avodagro's number - $6.022 \cdot {10}^{23}$ atoms in 1 mole - to determine how many atoms of nickel you have

$\text{0.0170 moles" * (6.022 * 10^(23)"atoms")/"1 mole" = 1.02 * 10^(22)"atoms of Ni}$

Now for the lithium aluminium hydride (LAH), or $L i A l {H}_{4}$. Notice that 1 molecule of LAH has 1 atom of lithium, 1 of aluminium, and 4 of hydrogen. Determine how many molecules of LAH you have first, then go to its component atoms.

$\text{0.10 g" * "1 mole"/"38.0 g" = "0.00263 moles LAH}$

$\text{0.00263 moles" * (6.022 * 10^(23)"molecules")/"1 mole" = 1.6 * 10^(21)"molecules of LAH}$

The number of atoms of Li, Al, and H will be

$1.6 \cdot {10}^{21} \text{molecules LAH" * "1 atom Li"/"1 molecule LAH" = 1.6 * 10^(21)"atoms Li}$

$1.6 \cdot {10}^{21} \text{molecules LAH" * "1 atom Al"/"1 molecule LAH" = 1.6 * 10^(21)"atoms Al}$

$1.6 \cdot {10}^{21} \text{molecules LAH" * "4 atoms H"/"1 molecule LAH" = 6.4 * 10^(21)"atoms H}$

Now for the oxygen. Gases are a little trickier because you need values for temperature and pressure to determine the number of moles. Since no mention of particular conditions was made, I'll assume STP - 273.15 K and 1 atm.

The great part about STP is that 1 mole of any ideal gas occupies exactly 22.4 L - this is called the molar volume of a gas. So if you know the volume, you can determine how many moles of oxygen you have.

Remember that air is only 20.9% oxygen, so the volume of oxygen gas will be

$\text{1 mL air" * "20.9 mL oxygen"/"100 mL air" = "0.209 mL }$ ${O}_{2}$

Another thing to look out for - notice that you've got diatomic oxygen, or ${O}_{2}$. This means that the number of oxygen atoms will be twice the number of ${O}_{2}$ molecules. So,

n = V/V_("molar") = (0.209 * 10^(-3)"L")/"22.4 L" = 9.3 * 10^(-6)"moles "# ${O}_{2}$

$9.3 \cdot {10}^{- 6} \text{moles" * (6.022 * 10^(23)"molecules" O_2)/("1 mole"O_2) = 5.6 * 10^(18)"molecules oxygen}$

Therefore, the number of oxygen atoms will be

$5.6 \cdot {10}^{18} \text{molecules oxygen" * "2 oxygen atoms"/"1 oxygen molecule" = 1.1 * 10^(19)"atoms of oxygen}$

You can use the same strategy to calculate how many atoms of nitrogen you have. Keep in mind that it too is a diatomic molecule, ${N}_{2}$, and that it's percentage in air's composition is 78.1%.