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Answer 1 · 2016-02-29T12:47:28

A) $6.0*10^22$ molecules of $CO_2$
B) $6.0*10^22$ molecules $O_2$

You are given 4.4g of $CO_2$ gas. To calculate the molecules of $CO_2$ present, you need to first covert from grams of $CO_2$ to moles of $CO_2$ .

4.4g $CO_2$ * $(1 mol)/(44.1g)$ = .10 mol of $CO_2$

After you covert to moles of $CO_2$ , you then covert to molecules of $CO_2$ with Avogadro's number; $6.022 * 10^23$ ,

.1 mol $CO_2$ $(6.022*10^23)/(1 mol)$ = $6.0*10^22$ molecules of $CO_2$

to do all your conversion in one step;

4.4g $CO_2$ $(1mol)/(44.1g)$ $(6.022*10^23)/(1mol)$ = $6.0*10^22$ molecules of $CO_2$

To determine the atoms of oxygen present;

You are given 4.4g $CO_2$ , but this time we want atoms (or molecules) of Oxygen gas.

4.4g $CO_2$ * $(1mol)/(44.1g)$ = .10 mol $CO_2$

For every mol of $CO_2$ , there is 1 mol of oxygen $(O_2)$

.10 mol $CO_2$ $(1mol O_2)/(1mol CO_2)$ =.10 mol $O_2$

we use Avogadro's number again;

.10 mol $O_2$ $(6.022*10^23)/(1mol O_2)$ = $6.0*10^22$ molecules/atoms $O_2$

All worked out;

4.4g $CO_2$ $(1mol CO_2)/(44.1g CO_2)$ $(1mol O_2)/(1mol CO_2)$ $(6.022*10^23)/(1mol O_2)$ = $6.0*10^22$ molecules of $O_2$