# What molar quantity is associated with a 0.44*g mass of "carbon dioxide"?

##### 1 Answer
Jul 29, 2017

We solve the quotient.......$\text{mass"/"molar mass}$ to get the number of moles........

#### Explanation:

And so here.........we gets..........$\frac{0.44 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1} = 0.010 \cdot m o l$ with respect to $C {O}_{2}$.

But one mole specifies by definition, ${N}_{A} , \text{Avogadro's number}$ of molecules, where ${N}_{A} \equiv 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And thus number of individual $C {O}_{2}$ molecules is given by the product..$6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \times 0.010 \cdot m o l \equiv 6.022 \times {10}^{21}$ individual molecules of $C {O}_{2}$. How many carbon atoms, and how many oxygen atoms does this molar quantity represent?