A compound made up of C, H, and Cl contains 55,0 percent Cl by mass. If 9,00 g of the compound contain 4,19 x 10^23 H atoms, what is the empirical formula of the compound?

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  1. I would do this.
    g Cl = 9.00 x 0.55 = ??
    g H = (4.19E23/6.02E23)*atomic mass H = ??
    g C = 9.00-gC-gH.

    Then convert g of each to mole of each and find the ratio to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself, then divide the other two numbers by the same small number. Round the final numbers to whole numbers. C2H5Cl is the probably answer but you need to confirm that. I worked some of this in my head.

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