Question

Glucose (C6H12O6) is a key nutrient for generating chemical potential energy in biological systems. We were provided 16.55 g of glucose. Please calculate:

a) The mass percent of carbon in glucose.

b) The mass of CO2 produced by the combustion of 16.55 g glucose with sufficient oxygen gas.

c) How many oxygen molecules needed for the completely combustion of 16.55 g glucose?

Answer 1

a) 40 %

b)
`4.04~g~CO_2`

c)
`5.53x10^2^3~molecules~of~O_2`

Step-by-step explanation:

For a) we will have to calculate the molar mass of
`C_6H_1_2O_6`, so the first step is to find the atomic mass of each atom and multiply by the amount of atoms in the molecule.

C => 12*(6) = 72

H => 1*(12) = 12

O => 6*(16) = 96

Molar mass = 180 g/mol

Then we can calculate the percentage by mass:

`Percentage~=~(72)/(180)*100=40`

For b) we have to start with the reaction of glucose:

`C_6H_1_2O_6~+~6O_2~->~6CO_2~+~6H_2O`

Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol carbon dioxide

-) 1 mol carbon dioxide = 44 g carbon dioxide

`16.55~g~C_6H_1_2O_6(1~mol~C_6H_1_2O_6)/(180~g~C_6H_1_2O_6)(6~mol~CO_2)/(1~mol~C_6H_1_2O_6)(44~g~CO_2)/(1~mol~CO_2)=4.04~g~CO_2`

For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol oxygen

-) 1 mol oxygen = 6.023x10^23 molecules of O2

`16.55~g~C_6H_1_2O_6(1~mol~C_6H_1_2O_6)/(180~g~C_6H_1_2O_6)(6~mol~O_2)/(1~mol~C_6H_1_2O_6)(6.023x10^2^3~molecules~O_2)/(1~mol~O_2)=~5.53x10^2^3~molecules~of~O_2`