# Determine the formula of the substance which has: 43.39%Na, 11.32%C, 45.28%O.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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Assuming that the %ge given in the question refer to the %ge weight of each element in the the molecule of the compound formed, The proportion of number atoms of each element in the molecule of the substance is given by %ge divided by atomic number.

The atomic weight of the three elements are:

Na: 22.9898

C : 12.011

O : 15.994

Therefore proportion of atoms of three elements are:

Na: 43.39/22.9898 = 1.887

C : 11.32/12.011 = 0.942469

O : 45.28/15.994 = 2.83106

Dividing these proportions by 0.942469 we get proportions of the three elements in whole numbers as follows:

Na: 1.887/0.942469 = 2

C :  0.942469/0.942469 =1

O :  2.83106/0.942469 = 3

From the above proportion we can conclude that the chemical formula of the substance is Na2CO3.

This is the formula of chemical compound sodium carbonate, which is also known as washing soda and soda ash.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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First of all, let's establish the formula for the substance as being:NaxCyOz, where

x=the number of Na atoms;

y= the number of C atoms;

z= the number of O atoms;

Then, we'll establish the atomic mass for each element, the notation used for the atomic mass being A:

A(Na)=23

A(C)=12

A(O)=16

Now, we'll calculate the number of gram-atoms of each element from molecule, by dividing the quantitative percentage to the atomic mass of the element. So:

x=43.39/23=1.89 gram-atoms of Na

y=11.32/12=0.94 gram-atoms of C

z=45.28/16=2.83 gram-atoms of O

The mixing up proportion of atoms:

x:y:z=1.89:0.94:2.83

We'll divide all obtained numbers to the smaller obtained, which is 0.94.

x:y:z=1.89/0.94:0.94/0.94:2.83/0.94

x:y:z= 2:1:3

The chemical formula of the substance is: Na2CO3

felando | Student, Undergraduate | (Level 2) eNoter

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Na=43.39%,  C=11.32%, O=45.28%

M.M.of Na=22.99,M.M.of C=12.01,M.M of O=16.00(M.M is molecular mass)

Divide the %  composition by its molecular mass

Na  =43.39/22.99  =1.89

C=11.32/12.01 =0.94

O=45.28/16  =2.83

Divide through by the smallest ratio,which is 0.94

Na=1.89/0.94=2.01

C=0.94/0.94=1.00

O=2.83/0.94=3.01

Then,write each symbol with the corresponding subscript

Na2CO3