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The molecular formula :
C₆H₁₄O₃PF
<h3>Further explanation</h3>
Given
39.10% carbon, 7.67% hydrogen, 26.11% oxygen, 16.82% phosphorus, and 10.30% fluorine.
Required
The molecular formula
Solution
mol ratio :
C = 39.1 : 12 = 3.258
H = 7.67 : 1 = 7.67
O = 26.11 : 16 = 1.632
P = 16.82 : 31 = 0.543
F = 10.3 : 19 = 0.542
Divide by 0.542
C = 6
H : 14
O = 3
P = 1
F = 1
The empirical formula :
C₆H₁₄O₃PF
(The empirical formula)n = the molecular formula
(C₆H₁₄O₃PF)=184.1
(6.12+14.1+3.16+31+19)n=184.1
(184)n=184.1
n = 1
Answer:
690 g IrI₃
Explanation:
To convert from moles to grams, you have to use the molar mass of the compound. The molar mass of IrI₃ is 572.92 g/mol. You use this as the unit converter in this equation:
Round to the lowest number of significant figures which is 2 to get 690 g IrI₃.
