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A compound is found to contain 2.270 % hydrogen, 34.80 % phosphorus, and 62.93 % oxygen by mass. What is the empirical formula for this compound

Sagot :

Answer: The empirical formula is [tex]H_4P_2O_7[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of H= 2.270 g

Mass of P = 34.80 g

Mass of O = 62.93 g

Step 1 : convert given masses into moles

Moles of H =  [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{2.270g}{1g/mol}=2.270[/tex]

Moles of P =  [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{34.80g}{31g/mol}=1.122[/tex]

Moles of O = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{62.93g}{16g/mol}=3.933[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For H =  [tex]\frac{2.270}{1.122}=2[/tex]

For P=    [tex]\frac{1.122}{1.122}=1[/tex]

For O =   [tex]\frac{3.933}{1.122}=3.5[/tex]

The simples ratio will be = H: P : O= 4: 2: 7

Hence the empirical formula is [tex]H_4P_2O_7[/tex]