empirical formula help?

Question

there are two common oxides of sulfur. One contains 32g of sulfur for each 32g of oxygen. The other oxide contains 32g of sulfur for each 48g of oxygen. What are the empirical formulas for the two oxides?

Answer 1

1) To determine the two empirical formulae we need to know the molecular masses of all chemical species:

M of S = 32 g/mol
M of O = 16 g/mol

2) We have to compute the number of moles of each species in the samples:

Sample 1: n of S = 32 g/32 g/mol = 1 mole
...................n of O = 32 g/16 g/mol = 2 moles

Ratio of moles is : n(O) / n(S) = 2/1 = 2 moles of oxygen per 1 mol of sulphur. Hence, empirical fromula of first oxide is:

SO2

Sample 2: n of S = 32 g/32 g/mol = 1 mole
..................n of O = 48g/16 g/mol = 3 moles

Ratio of moles is: n(O)/n(S) = 3/1 = 3 moles of oxygen per 1 mol of sulphur. Hence, the empirical formula of second oxide is:

SO3

That´s it!

Good luck!

Answer 2

First:

SO

Because if there is 32 g of sulfur for each 32g of oxygen, it means that there is 1 atom of sulfur for each one of oxygen (32 sulfur/32oxygen=1sulfur/1oxygen), so it can be S1O1, S2O2, S3O3, etc. Simplfied, they will all mean S1O1, which means SO

Second:

SO2

Same as first! 32/48=2sulfur/3oxygen so the formula would be S2O3.

Answer 3

SO2 1 sulphur=32g/mole; 2 O @16g/mole=32g

SO3 1 S=32g/mole; 3 O at 16g/mole=48g

Answer 4

Empirical formulation teach the relative numbers of atoms in a molecular compound. subsequently they are type of like the "lowest difficulty-loose denominator" in a math challenge. The molecular formulation is the certainly, not relative, type of atoms in a molecular compound. in spite of the indisputable fact that, often times, with straight forward compounds, the molecular formulation might want to be a twin of the empirical formulation. Examples; a million) octane = C8H18. Thats that is molecular formulation, and that is the precise type of each and every atom in a unmarried molecule of octane. the empirical formulation may be the bottom entire volume ratio of 8 and 18 (ie: like the "lowest difficulty-loose denominator" theory). Divide each and each and every by technique of two, and also you are able to get C4H9, and also you may't decrease that any more effective and nonetheless have entire numbers, so C4H9 is the empirical formulation. 2) methane = CH4 that is the molecular formulation for methane. It can't be decreased any farther, so it really is likewise the empirical formulation (by technique of definition). be conscious: something with in basic terms one molecule of ANY atom is going to be the empirical formulation AND the molecular formulation, as you won't be able to diminish ONE atom to something less than one, and nonetheless be an finished volume. throughout the time of experiments, the ratio of factors to at least one yet another might want to be stumbled on particularly genuinely. subsequently, you'll locate the empirical formulation. to substantiate the molecular formulation from that information, you want to locate the molecular weight of the compound, also extremely undemanding. the certainly molecular formulation will be an finished volume diverse of the empirical formulation, (and understand that the entire volume might want to be in basic terms "one", not inevitably 2, 3, 4 and so on.).

Answer 5

Consult that dusty, lonely textbook of yours and determine the mass of O2 and do your own homework.