a compound has a percent composition of 42.1% carbon, 6.5% hydrogen, and 51.4% oxygen. the compound's total mass is 342.30008 amu. what is the formula for this compound?
Atomic mass units of mentioned elements are the following: Carbon – 12.011 a. m. u. Oxygen – 15.999 a. m. u. Hydrogen – 1.0079 a. m. u.
Firstly we must determine the total mass of atoms of each element of compound.
The total mass of atoms of Carbon will be calculated in the following way: 42.1%/100%*342.30008 a. m. u. = 144.108 a. m. u. So, the number of atoms of Carbon will be: 144.108/12.011 = 12 For Oxygen this calculation will have the following view: 51.4%/100%* 342.30008 / 15.999 = 11. Finaly for Hydrogen: 6.5%/100%* 342.30008 / 1.0079 = 22. So, the empirical formula of this compound is C12H22O11.
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