Question

if the percentage of sodium is equal to 42.07 and percentage of phosphorus is 18.9 and the percentage of oxygen is 39.03
what is the empirical formula​

Answer 1

Answer:

hi

Explanation:

We gots #Na_3PO_4#

Explanation:

As is customary with these problems, we assume a #100*g# mass of compound, and then we interrogate its atomic composition...

#"Moles of sodium"-=(42.07*g)/(22.99*g*mol^-1)=1.83*mol#

#"Moles of phosphorus"-=(18.89*g)/(31.00*g*mol^-1)=0.590*mol#

#"Moles of oxygen"-=(39.04*g)/(16.00*g*mol^-1)=2.44*mol#

And then we normalize the result by dividing thru by the atom in the LEAST molar quantity...i.e. #P#, to give an empirical formula of...

#Na_((1.83*mol)/(0.590*mol))O_((2.44*mol)/(0.590*mol))P_((0.590*mol)/(0.590*mol))-=Na_3.1O_4.14P#..the which is close enuff to #Na_3PO_4#..#"natrium phosphate"#