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What is the molecular formula of a compound that has an empirical formula of c2oh4 and a molar mass of 88 grams per mole?
The empirical formula C2OH4 simplifies to C1O2H2. To find the molecular formula, divide the molar mass (88 g/mol) by the empirical formula mass (112 + 216 + 2*1 = 46 g/mol) to get 1.91. This means the molecular formula is approximately 1.91 times the empirical formula, so the molecular formula is C2O4H4.
What is the molecular formula of a compound with an empirical formula of C2OH4 and a molar mass of 88 grams per mole?
Look at your atomic masses for phosphorus and oxygen. Set up 2.5/31 and 3.23/16. This makes 0.08 and 0.2. Now consider 0.08/0.2 to equal 0.4. So, your ratio of P to O is 0.4, which is 4/10. So, your formula, which reflects that ratio, is P4O10. That's tetraphosphorus decoxide, a molecular solid used in match-heads.
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