Question

2.2g of an inorganic salt contains 0.954g of sodium, 0.25g of carbon and the rest is oxygen. Calculate its Empirical formula.

Answer 1

Answer:

Here is your answer buddy

Answer 2

Answer:

So, the Empirical formula will be = Na₂CO₃

Explanation:

It is given that :

total amount of salt given = 2.2g

For sodium (Na),

mass given = 0.954g

Therefore percentage of the mass of sodium = $\frac{0.954}{2.2} * 100$ % = 43.36 %

For carbon(C),

mass given = 0.25g

Therefore percentage of the mass of carbon = $\frac{0.25}{2.2} * 100$ % = 11.36 %

For oxygen (O),

mass = 2.2 - (0.954 + 0.25) = 0.996g

Therefore percentage of the mass of oxygen = $\frac{0.996}{2.2} *100$ % = 45.27 %

table :

ELEMENT   PERCENTAGE  ATOMIC MASS    NO. OF ATOMS   SIMPLEST            

                                                                                                         FORM

Na                     43.36%               23                  43.36/23=1.88   1.88/0.94=2

C                        11.36 %               12                  11.36/12=0.94     0.94/0.94=1

O                        45.27%               16                 45.27/16=2.83   2.83/0.94=3

So, the Empirical formula will be = Na₂CO₃

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