How do i find the empirical formula of a compound that is 35.3% carbon, 8.82% hydrogen, and 55.88% fluorine?

you divide each % by the molar mass of the element 35.3/12=2.93 approximately 3 8.82/1=8.82 approximately 9 55.88/19=2.94 approximately 3 so you have as much flurine and cerbon and 3 times more Hydrogen results CH3F

Assume that there are 100 grams of this compound which would mean 35.3g C, 8.82g H and 55.88g F. Convert these grams into moles by multiplying them by its molar mass. (35.3 g C)/(12.01 grams/mol C) = 2.94 moles of C (8.82 g H)/(1.008 grams/mol H) = 8.75 moles of H (55.88 g F)/(19.00 grams/mol F) = 2.94 moles of F Divide each mole of each atom by its smallest mole ratio (2.94 moles of C)/(2.94) = 1 C (8.75 moles of H)/(2.94) is approximately 3 H (2.94 moles of F)/(2.94) = 1 F The empirical formula = CH3F

1. Start with the assumption that you have 100g of the compound. That means you have 35.3g C, 8.82g H and 55.88g F. 2. Change each to moles using MM. 3. Divide the smallest number of moles into the the two larger to determine mole ratio. Write the ratio out in alphabetical order, ie C1H2O1 4. If any fraction of a mole, multiply the whole compound to make smallest whole number ratio. EX: C0.25H1 multiply by 4 to get C1H4

you divide each % by the molar mass of the element 35.3/12=2.93 approximately 3 8.82/1=8.82 approximately 9 55.88/19=2.94 approximately 3 so you have as much flurine and cerbon and 3 times more Hydrogen results CH3F

Assume that there are 100 grams of this compound which would mean 35.3g C, 8.82g H and 55.88g F. Convert these grams into moles by multiplying them by its molar mass. (35.3 g C)/(12.01 grams/mol C) = 2.94 moles of C (8.82 g H)/(1.008 grams/mol H) = 8.75 moles of H (55.88 g F)/(19.00 grams/mol F) = 2.94 moles of F Divide each mole of each atom by its smallest mole ratio (2.94 moles of C)/(2.94) = 1 C (8.75 moles of H)/(2.94) is approximately 3 H (2.94 moles of F)/(2.94) = 1 F The empirical formula = CH3F