A sample of chrominum oxide contains 52% Cr and the rest O. Write the simplest formula and name the compound.

Question

Use these atomic masses

Cr = 52.00 g/mol

O = 16.00 g/mol

Moles of Cr = 52 g / 52.00 g/mol = 1.000 mol Cr

Moles of O = 48 g / 16.00 g/mol = 3.000 mol O

1 / 1 = 1 Cr : 3 / 1 = 3 O

The empirical formula is CrO3, also known as

chromium (III) oxide. Is the name correct? Check.

Answer 1

Chromium(VI) trioxide.

Bye

Answer 2

Yes, very well done.

Answer 3

correct and easy Cr has an atomic mass of 52 so 52% =52

!% =1

oxygen represents 48% so mass 48g =O3 (3*16)

fromula CrO3 chromium III oxide