The mass percent of oxygen in pure glucose, c6h12o6 is 53.3 percent. a chemist analyzes a sample of glucose that contains impurities and determines that the mass percent of oxygen is 49.7 percent. which of the follow impurities could account for the low mass percent of oxygen in the sample?

We can calculate the mass percent of an element by dividing its atomic mass by the mass of the compound and then multiply by 100:
     % by mass of element = (mass of element/mass of compound) x100%
Impurities like n-eicosane with the molecular formula C20H42 could account for the low percent by mass of oxygen in the sample because it has a zero percent oxygen based on its compound formula which indicates that it does not have the element oxygen.

If it releases 5 tons a day and there are 7 days a take that answer and multiply by 5 percent (0.05) to find the amount released in one week. leave a comment if you need more .