Final answer:
The empirical formula of glucose, based on its mass percent composition of C 40.00%, H 6.71%, O 53.28%, is CH₂O. The mass percent composition is used to find the simplest whole-number ratio of the elements in the compound by converting the masses of each element to moles and then finding the ratio of the moles. In this case, the ratio is 1:2:1 for carbon, hydrogen, and oxygen, respectively.
Step-by-step explanation:
The empirical formula of a compound can be determined using its mass percent composition. In the case of glucose, the mass percent composition is: C 40.00%, H 6.71%, O 53.28%. To find the empirical formula, we need to determine the simplest whole-number ratio of the elements in the compound. Starting with the mass percent composition, we assume a 100g sample of the compound to make the calculations easier. This means there would be 40.00g of carbon, 6.71g of hydrogen, and 53.28g of oxygen.
Next, we need to convert the masses of each element to moles. The molar mass of carbon is 12.01 g/mol, hydrogen is 1.01 g/mol, and oxygen is 16.00 g/mol. Using these values, we divide the mass of each element by its molar mass to get the number of moles. For carbon, it would be 40.00g / 12.01 g/mol = 3.33 mol. For hydrogen, it would be 6.71g / 1.01 g/mol = 6.65 mol. And for oxygen, it would be 53.28g / 16.00 g/mol = 3.33 mol.
Finally, we divide the number of moles of each element by the smallest number of moles to get the simplest ratio. In this case, the smallest number of moles is 3.33 mol. Dividing the moles of carbon and oxygen by 3.33 gives us a ratio of 1:1. And dividing the moles of hydrogen by 3.33 gives us a ratio of 2:1. Therefore, the empirical formula of glucose is CH₂O.