Question ee296

Apr 25, 2015

Starting with the first one

$\textcolor{b l u e}{\left(1\right)}$ One mole of any substance contains exactly $6.022 \cdot {10}^{23}$ molecules or atoms of that substance - this is known as Avogadro's number. Since you have less than 1 mole of ethanol, $C {H}_{3} O H$, you'll get less that $6.022 \cdot {10}^{23}$ molecules of ethanol.

$3.10 \cdot {10}^{- 2} \cancel{\text{moles") * (6.022 * 10^(23)"molec.")/(1cancel("mole")) = color(green)(1.87 * 10^(22)"molec.}}$

$\textcolor{b l u e}{\left(2\right)}$ To get the number of carbon atoms present in that much butane, you must first determine how many molecules of butane you're dealing with. Since the molecular formula of butane is ${C}_{4} {H}_{10}$, you'll get 4 carbon atoms for every 1 molecule of butane.

0.570cancel("moles") * (6.022 * 10^(23)"molec.")/(1cancel("mole")) = 3.43 * 10^(23)"molec."

Therefore, you get

3.43 * 10^(23)cancel("molecules") * ("4 C atoms")/(1cancel("molecule")) = color(green)(1.37 * 10^(24)"atoms"#

Questions $\textcolor{b l u e}{\left(3\right)}$ and $\textcolor{b l u e}{\left(4\right)}$ are identical to the first two, the only difference being the actual values you have to work with.

For question $\textcolor{b l u e}{\left(4\right)}$, the number of atoms of oxygen in aluminium nitrate will be 9. You get 3 oxygen atoms from each of the 3 nitrate anions that make up the compound.