Question #211582

Note:

1 mole of a compound is equivalent to its molecular mass

1 mole of a compound is equal to 6.02x10^23 molecules

The percentage is a direct equivalent to grams

Answers should be in 3 significant figures

No need to include the units and the compound in the answers


  1. A compound is found to contain 63.52 % iron and 36.48% sulfur. Find its empirical formula. 
  2.   A compound is found to contain 26.56% potassium, 35.41% chromium, and the remainder oxygen. Find its empirical formula.  

 

  




1
Expert's answer
2021-06-29T00:38:37-0400

1. One gram of a compound contain 0.6352 g iron and 0.3648 g sulfur.

n(Fe)=0.635255.84=0.011375  moln(S)=0.364832.06=0.011378  moln(Fe) = \frac{0.6352}{55.84} = 0.011375 \; mol \\ n(S) = \frac{0.3648}{32.06} = 0.011378 \; mol

Proportion:

n(Fe) : n(S) = 0.011375 : 0.011378 = 1:1

Empirical formula: FeS

2. One gram of a compound contain 0.2656 g potassium and 0.3541 g chromium, and 0.3803 g oxygen.

n(K)=0.265639.10=0.006792  moln(Cr)=0.354151.99=0.006810  moln(O)=0.380315.99=0.023783  moln(K) = \frac{0.2656}{39.10}=0.006792 \;mol \\ n(Cr) = \frac{0.3541}{51.99}= 0.006810 \;mol \\ n(O) = \frac{0.3803}{15.99}= 0.023783 \;mol

Proportion:

n(K) : n(Cr) : n(O) = 0.006792 : 0.006810 : 0.023783

=0.0067920.006792:0.0068100.006792:0.0237830.006792=1:1:3.5=2:2:7= \frac{0.006792}{0.006792} : \frac{0.006810}{0.006792} : \frac{0.023783}{0.006792} \\ = 1 : 1 : 3.5 = 2: 2: 7

Empirical formula: K2Cr2O7

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