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The best way of doing this is by invoking the concept of moles and avagadro's number. Avagadros number is 6.022*10^23. One mole is an avagadro's number of atoms. So to solve this problem a Periodic Table can be used to find the molar mass (the mass of one mole) of C6H12O6. This comes out to be: (6*12.01)+(12*1)+(6*16)=180.06 g/mole.

Then we use dimensional analysis to find the number of C, H, and O atoms:

Carbon atoms:

3.00g*(1 mole C6H12O6/180.06g)*(6 mole C/1 mole C6H12O6)*(6.022*10^23 atoms /mole C)=6.02*10^22 atoms of carbon

Hydrogen atoms:

3.00g*(1 mole C6H12O6/180.06g)*(12 mole H/1 mole C6H12O6)*(6.022*10^23 atoms /mole H)=1.20*10^23 atoms of hydrogen

Oxygen atoms:

3.00g*(1 mole C6H12O6/180.06g)*(6 mole H/1 mole C6H12O6)*(6.022*10^23 atoms /mole H)=6.02*10^22 atoms of oxygen

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15y ago
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10y ago

Glucose is C6H12O6 so 3 molecules would contain 18 Cs , 36 Hs, and 18 Os.

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10y ago

There are 18.

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Q: How many carbon oxygen and hydrogen atoms will you need for three glucose molecules?
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