An element X forms two oxides, Formula of first oxide is XO2.The first oxide contains 50% of oxygen.If the second oxide contains 60% oxygen ,the formula of the second oxide is_____

Yes u are correct.

 

Percent of oxygen in first oxide = 50%

Percent of metal in first oxide = 50%

Let the atomic mass of the metal be m g

Molecular mass of first oxide ‘XO2’ = m + 16x2 = (m+32)g

Percentage of oxygen by mass in the first oxide = 32/(m+32)*100 = 50

On solving mass of metal m = 32 g

 

Determination of the formula of second oxide

Element  %  At. mass  relative no.of atoms  simple ratio   simplest whole no. ratio 

1. Metal  40   32   40/32 = 1.25  1.25/1.25 =1.0   1.0 x1 =1

2. Oxygen 60  16  60/16 = 3.75  3.75/1.25 = 3.0  3.0x1 =3

So the formula of second oxide is XO3

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Let the mass of metal be M

if Ist oxide contain 50% of oxygen, then

                    32/(M+32)=50/100

             M=32

NOW,     for 2nd oxide-

   60% of oxygen is present then metal would be 40%

so,     40/M : 60/16

            40/32 : 60/16 = 1.25 : 3.75 =1:3

        2nd oxide is XO3

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