Problem: Glucose (a source of energy and metabolic intermediate): 40.00% C, 6.71% H, and 53.29% O.Express your answer as a chemical formula.

FREE Expert Solution

Recall that the empirical formula is the simplest whole-number ratio formula of a compound


Step 1: Recall that mass percent is given by:


% mass=mass of Xtotal mass×100


Assuming we have 100 g of the compound, this means we have 40.0 g C, 6.71 g H, and 53.29 g O.



Now, we need to get the moles of each element in the compound. 

The atomic masses are 12.01 g/mol C, 1.01 g/mol H, and 16 g/mol O.


40.0 g C×1 mol C12.01 g C = 3.33 mol C


6.71 g H×1 mol H1.01 g H = 6.64 mol H


53.29 g O×1 mol O16.00 g O = 3.33 mol O


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Problem Details

Glucose (a source of energy and metabolic intermediate): 40.00% C, 6.71% H, and 53.29% O.

Express your answer as a chemical formula.

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