# How many ATOMS in a 18*g mass of glucose?

Jul 1, 2017

Approx. $\frac{1}{10} \cdot m o l \times 24 \cdot \text{atoms} \cdot m o {l}^{-} 1 \times {N}_{A}$,

where ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

#### Explanation:

We take the quotient.....

"Number of moles"=("Mass of substance"*g)/("Molar mass"*g*mol^-1)

$= \left(\text{Mass of substance"*cancelg)/("Molar mass} \cdot \cancel{g} \cdot m o {l}^{-} 1\right) = \frac{1}{\frac{1}{m o l}} = m o l$

And thus we get an answer with dimensions of $m o l$ as we require...

And so we retake the quotient.......

"Number of moles of glucose"=(18*g)/(180.16*g*mol^-1)=??*mol.

We know that glucose, ${C}_{6} {H}_{12} {O}_{6}$, has a formula of $180.16 \cdot g \cdot m o {l}^{-} 1$, i.e. $\left(6 \times 12.011 + 12 \times 1.00794 + 6 \times 16.00\right) \cdot g \cdot m o {l}^{-} 1$

=??*g*mol^-1......

We are not finished there, because we were asked to find the number of atoms in such a molar quantity, and there are 24 atoms in one molecule of glucose.

And so we multiply the molar quantity by number of atoms per mole, and then by Avogadro's number to get the number of atoms, $\frac{18 \cdot g}{180.16 \cdot g \cdot m o {l}^{-} 1} \times 24 \cdot \text{atoms} \cdot m o {l}^{-} 1 \times {N}_{A}$

(18*g)/(180.16*g*mol^-1)xx24*"atoms"*mol^-1xx6.022xx10^23*mol^-1=??

Jul 1, 2017

There are $6.0 \times {10}^{22}$ atoms of glucose in $\text{18 g}$ of glucose.

#### Explanation:

One mole of anything is $6.022 \times {10}^{23}$ of them, including atoms. The mass of the compound is converted into moles by dividing the given mass by its molar mass. Once you have moles, you multiply by $6.022 \times {10}^{23} \text{atoms/mol}$.

The molar mass of glucose is $\text{180.156 g/mol}$.
https://www.ncbi.nlm.nih.gov/pccompound?term=Glucose

Divide the given mass of glucose by its molar mass by inverting the molar mass (a fraction) and multiply.

18color(red)cancel(color(black)("g C"_6"H"_12"O"_6))xx("mol C"_6"H"_12"O"_6)/(180.156color(red)cancel(color(black)("g C"_6"H"_12"O"_6)))="0.10 mol C"_6"H"_12"O"_6"

Multiply mol glucose by $\left(6.022 \times {10}^{23} {\text{atoms C"_6"H"_12"O"_6)/(1"mol C"_6"H"_12"O}}_{6}\right)$.

color(red)cancel(color(black)(0.10"mol C"_6"H"_12"O"_6))xx(6.022xx10^23"atoms C"_6"H"_12"O"_6)/(1color(red)cancel(color(black)(1"mol C"_6"H"_12"O"_6)))=6.0xx^22 "atoms C"_6"H"_12"O"_6" rounded to two sig figs