[sjb]

How many moles of M?

[IBM]

1 mole of M

[sjb]

No, if there are for example 2 moles of O in a sample of MO, how many moles of M are there?

[IBM]

2 moles of M are there
2M + 2O2 = 2MO

[sjb]

So in a sample of MO with 1.725 mole of O, how much M is there? (Ignoring the improperly balanced equation as that's not relevant at the moment)

[IBM]

1.725 moles of M weighs 74.4g
So, 1 mole of M weighs 74.4/1.725 = 43.13g 

[sjb]

Good, so in 100g of your second sample containing 30.0 % O, what is the mass of M, how many moles of M and O do you have?

[IBM]

70g of M,
1.875 moles of O and M

[sjb]

No, if the sample is 70g of M, with molecular weight 43.13g mol^-1, what is the number of moles? (Does it make sense for both numbers of moles to be increased from your first sample?)

[IBM]

43.13g = 1 mole So, 70g = (70/43.13) = 1.62 moles

[sjb]

So what is the ratio in your second sample?

[IBM]

But how to calculate the moles' of O in the second sample?
would it be 43.13g O = 1 mole
               So, 30g O =.69 mole
So. the ratio in the second sample is M:O = 1.62:.69 = 54:23

[sjb]

You knew the mass of oxygen for sample 2 right at the start, and the atomic mass of oxygen is readily available.

[IBM]

So, what to do now?

[IBM]

Two oxides of a metal contain 27.6% and 30% of oxygen respectively. The formula is MO.
Now my questions are below:
In the case of the first formula MO:
1. If the compound is 'MO' as a product, The reactant would be 'M +0.5 O2 = MO'
So if the total amount is 100 g (MO), M + 0.5 O2 =100g. From this equation in the left side, what will be the mass of O and M (not from the MO because I am able to calculate the mass of O and M from 'MO')?