With that in mind, calculate the percent composition of oxygen in C_12H_22O_11 (

Question

With that in mind, calculate the percent composition of oxygen in C_12H_22O_11 (molar mass = 342.34 g/mol). 4.67% 51.4% 24.4% 0.514% 0.421% The empirical formula is the simplest whole-number ratio of elements in a molecule. The molecular formula is essentially the actual count of all elements in the molecular. The molecular formula for a particular substance will always be some multiple of the empirical formula. What is the empirical formula for C_12H_18? (Please note that subscripts cannot be entered into the answer choices for some reason.) C12 H18 C6 H9 C4H6

Explanation / Answer

Given

compound is C12H22O11

Molecular weight of given compound is 342.34 g/mol

let us consider 1 mol of compound

it will weigh 342.34 g

Molecular weight of O = 16 g/mol

there are 11 moles of O in 1 mole of compound

so weight of O in 1 mole of compound = 11 * 16 = 176 g

percent of O in compound = (176 g / 342.34 g )* 100 % = 51.4 % answer

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