# How many atoms of carbon can be found in #"90 g"# of glucose?

##### 1 Answer

#### Answer:

#### Explanation:

In order to determine how many atoms of carbon you get in

how manymoleculesyou have in#90# grams of glucosehow many atoms of carbon you haveper moleculeof glucose

So, glucose's molecular formula is **every** molecule of glucose contains

six carbon atomstwelve hydrogen atomssix oxygen atoms

Now that you know how many atoms of carbon you get per molecule of glucose, all you need to figure out is how many molecules of glucose you get in that sample.

To do that, use glucose's **molar mass**, which tells you what the mass of **one mole** of glucose molecules is. Glucose has a molar mass of **one mole** of glucose molecules has a mass of

Therefore, your

#90 color(red)(cancel(color(black)("g"))) * ("1 mole C"_6"H"_12"O"_6)/(180.156color(red)(cancel(color(black)("g")))) = "0.4996 moles C"_5"H"_12"O"_6#

As you know, the number of molecules you get *per mole* is given by **Avogadro's number**. More specifically, one mole of any substance contains exactly

This means that you have

#0.4996 color(red)(cancel(color(black)("moles"))) * (6.022 * 10^(23)"molecules")/(1color(red)(cancel(color(black)("mole")))) = 3.009 * 10^(23)"molecules"#

So, if you know how many molecules of glucose you have, and how many atoms of carbon you have per molecule, you can say that

#3.009 * 10^(23)color(red)(cancel(color(black)("molecules C"_6"H"_12"O"_6))) * "6 atoms of C"/(1color(red)(cancel(color(black)("molecule C"_6"H"_12"O"_6)))) = 1.81 * 10^(24)"atoms of C"#

You should round this off to one sig fig, the number of sig figs you have for the mass of glucose

#"no. of atoms of C" = 2 * 10^(24)#