Question

a compound is 42% sodium, 19% phosphorus, and 39% oxygen by mass. what is the empirical formula of this compound?

Answer 1

`Na_(x)P_(y)O_(z)\\\\ 42g \ \ Na \Rightarrow \ \ \ (42g)/(23(g)/(mol))=1,8mol\\\\ 19g \ \ P \Rightarrow \ \ \ (19g)/(31(g)/(mol))=0,6mol\\\\ 39g \ \ O \Rightarrow \ \ \ (39g)/(16(g)/(mol))=2,4mol\\\\ Na_(1,8)P_(0,6)O_(2,4)\\\\\\ Na: \ \ (1,8)/(0,6)=3 \ \ \ P: \ \ (0,6)/(0,6)=1 \ \ \ O: \ \ (2,4)/(0,6)=4\\\\\\ Na_(3)PO_(4)`

Answer 2

Answer: The empirical formula of the compound is
`Na_3PO_4` .

Solution:- An empirical formula is a simplest whole number ratio of all the atoms present in a compound.

If we assume 100 grams for the mass of the compound then the given percentages of all the elements could be taken as their grams.

First of all we divide the grams of each elements by its atomic mass to get its moles.

`42gNa((1molNa)/(23gNa))` = 1.83 moles Na

`19gP((1molP)/(31gP))` = 0.613 moles P

`39gO((1molO)/(16gO))` = 2.44 moles O

Next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them.

If we look at the moles of each then least one of them is moles of P that is 0.613. So, let's divide the moles of each by 0.613.

`Na=(1.83)/(0.613)` = 3(round of value)

`P=(0.613)/(0.613)` = 1

`O=(2.44)/(0.613)` = 4(round of value)

So, the empirical formula is
`Na_3PO_4` .