# If mass % of Oxygen in a monovalent metal carbonate is 48. Then find the number of atoms of metal present in 5mg of this metal carbonate sample?( N_a = 6.022 * 10^23)

Sep 4, 2017

#### Answer:

$6.0 \times {10}^{19}$ $\text{atoms M}$

#### Explanation:

We're asked to find the number of atoms of the metal present in $5$ $\text{mg}$ of a monovalent metal carbonate sample.

A monovalent metal carbonate includes a metal with a valence of one; i.e. a metal in group 1 ($\text{Li}$, $\text{Na}$, $\text{K}$, etc.). The general formula will be

ul("M"_2"CO"_3

We can use dimensional analysis to find the number of moles of the metal, $\text{M}$, present in the compound:

5cancel("mg sample")((48cancel("mg O"))/(100cancel("mg sample")))((1cancel("g O"))/(10^3cancel("mg O")))((1cancel("mol O"))/(16.00cancel("g O")))((2color(white)(l)"mol M")/(3cancel("mol O"))) = color(red)(ul(0.0001color(white)(l)"mol M"

Now, we can use Avogadro's number, ${N}_{\text{A}}$ to find the number of atoms of $\text{M}$ in the sample:

$\textcolor{red}{0.0001} \cancel{\textcolor{red}{\text{mol M"))((6.022*10^23color(white)(l)"atoms M")/(1cancel("mol M"))) = color(blue)(ulbar(|stackrel(" ")(" "6.0 xx 10^19color(white)(l)"atoms M"" }} |}$

which I'll just round off to $2$ significant figures.

Therefore, there are color(blue)(6.0xx10^19color(white)(l)"atoms" of the metal present in the sample.

Sep 5, 2017

#### Answer:

$\textsf{6 \times {10}^{19}}$

#### Explanation:

The metal is monovalent so the formula can be written $\textsf{{M}_{2} C {O}_{3}}$.

The mass of oxygen in the sample is given by:

$\textsf{{m}_{O} = 0.48 \times 5 \times {10}^{- 3} = 2.4 \times {10}^{- 3} \textcolor{w h i t e}{x} g}$

To find the no. moles of oxygen we divide by the mass of 1 mole:

$\textsf{{n}_{O} = {m}_{O} / \left({A}_{r} \left[O\right]\right) = \frac{2.4 \times {10}^{- 3}}{16.0} = 0.15 \times {10}^{- 3}}$

From the formula we can see that the no. moles of $\textsf{M}$ must be 2/3 of this:

$\textsf{{n}_{M} = 0.15 \times {10}^{- 3} \times \frac{2}{3} = 0.0001 \textcolor{w h i t e}{x} \text{mol}}$

To find the no. of atoms of $\textsf{M}$ we multiply by The Avogadro Constant:

$\textsf{{n}_{M} = 6.022 \times {10}^{23} \times 0.0001 = 6.022 \times {10}^{19}}$