# Question a3a16

May 26, 2015

The unknown metal is iron, or $F e$.

So, you know that you get 2.327 g of your unknown metal for every 1.00 g of elemental oxygen. You can use these values to determine the percent composition of your oxide

${m}_{\text{oxide" = m_"metal" + m_"oxygen}}$

${m}_{\text{oxide" = 2.327 + 1.00 = "3.327 g}}$

This means that you get

(2.327cancel("g"))/(3.327cancel("g")) * 100 = "69.94%" $\to$ metal

(1.0cancel("g"))/(3.327cancel("g")) * 100 = "30.06%" $\to$ oxygen

To make the calculations easier, assume that you're dealing with a 100.0-g sample of ${M}_{\textcolor{red}{2}} {O}_{\textcolor{b l u e}{3}}$. You can determine how many moles of oxygen you'd get in this much oxide by using oxygen's molar mass.

Since oxygen has a percent composition of 30.06%, your sample will contain 30.06 g of oxygen. This means that you have

30.06cancel("g") * "1 mole oxygen"/(16.00cancel("g")) = "1.879 moles" $O$

Now use the known mole ratio that exists between the metal on oxygen to determine how many moles of metal the sample would have

1.879cancel("moles O") * (color(red)(2)"moles metal")/(color(blue)(3)cancel("moles O")) = "1.253 moles M"

Since the metal's percent composition in the oxide is equal to 69.94%, your sample will contain 69.94 g of metal. This means that its molar mass will be

M_M = m/n = "69.94 g"/"1.253 moles" = color(green)("55.8 g/mol")#

Your unknown metal is iron, $F e$, which has a molar mass of $\text{55.845 g/mol}$.