What is the empirical and molecular formula of the compound?

What is the empirical formula of the unknown compound?

  • Answer:

    mass of C in 17.7 g of CO2 = 17.7 x MM C / MM CO3 = 17.7 x 12 / 44 = 4.827 g mass of H = 7.26 x 2 / 18 = 0.8067 g mass of H + C = 5.6339 g so mass of O in the sample 7.8 - 5.6339 = 2.1661 g moles of C = 4.827 / 12 = 0.4023 moles moles of H = .8067 / 1.008 = 0.8003 moles moles of O = 2.1661 / 16 = 0.1354 moles molar ratio of C : H : O = 0.4023: 0.8003:: 0.1354 divide by the smallest number we get 3:6:1 empirical formula is C3H6O

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mass of C in 17.7 g of CO2 = 17.7 x MM C / MM CO3 = 17.7 x 12 / 44 = 4.827 g mass of H = 7.26 x 2 / 18 = 0.8067 g mass of H + C = 5.6339 g so mass of O in the sample 7.8 - 5.6339 = 2.1661 g moles of C = 4.827 / 12 = 0.4023 moles moles of H = .8067 / 1.008 = 0.8003 moles moles of O = 2.1661 / 16 = 0.1354 moles molar ratio of C : H : O = 0.4023: 0.8003:: 0.1354 divide by the smallest number we get 3:6:1 empirical formula is C3H6O

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