Your strategy here will be to use molar masses of the two elements to determine how many moles of each you have present in this sample.
Once you know that, use the number of moles of sulfur and oxygen to determine the smallest whole number ratio that exists between the two elements in the compound.
So, sulfur has a molar mass of #"32.065 g mol"^(-1)#, which means that one mole of sulfur has a mass of #"32.065 g"#.
Your sample will thus contain
#60.0 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "1.8712 moles S"#
Oxygen has molar mass of #"15.9994 g mol"^(-1)#, which means that one mole of oxygen will have a mass of #"15.9994 g"#.
Your sample will contain
#90.0 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "5.6525 moles O"#
Now, a compound's empirical formula tells you the smallest whole number ratio that exists between the compound's constituent elements.
To get the mole ratio that exists between sulfur and oxygen in this compound, divide both values by the smallest one
#"For S: " (1.8712 color(red)(cancel(color(black)("moles"))))/(1.8712color(red)(cancel(color(black)("moles")))) = 1#
#"For O: " (5.6525color(red)(cancel(color(black)("moles"))))/(1.8712color(red)(cancel(color(black)("moles")))) = 3.021 ~~ 3#
Since #1:3# is the smallest whole number ratio that can exist between the two elements, it follows that the compound's empirical formula will be
#color(green)(|bar(ul(color(white)(a/a)"S"_1"O"_3 implies "SO"_3color(white)(a/a)|)))#
Video Transcript
Problem. We have to find the empirical formula of a compound which is 40% sulfur and 60% oxygen by way. So let's write ratio of Ratio of Sulfur is to oxygen is 40, 40 is to 60 by mass or by weight. Okay, now Mueller masa Moeller, ma, sauve, Sal Fairies, 32 and moller maas off oxygen, smaller mass up oxygen is 16. So we can find out that um number of moles of sulfur and number of moles of oxygen. Therefore number of moon up number of moles of sulfur. It will be for D Divided by 32. There's the Mueller mass of suffer And there's actually atomic mass of Sulfur. So we'll have 1.25 and number of mall of oxygen oxygen 60 Moss divided by a smaller mass Or the atomic mass 16 and we get 3.75. So the molar ratio molar ratio of selfish to oxygen is 1.25 is 23.75, that is one is to three. So the empirical formula will be I am pretty good formula will be one sulfur, so S. And three for each cell for three oxygen atoms. So S. 03. So the option. Third option is the correct one. Third option is the correct answer. Empirical formula is 0. 3
Video Transcript
Here in this problem, we have to find out the empirical formula of a compound containing 60.0 percent sulphur and 40.0 percent oxygen by mass, so this compound contains 60 percent sulphur and 4540 percent oxygen in the sample is 100 gram. So 100 gram sample contains 60 grams. Sulphur and 40 gram oxygen now we'll find out moles of sulphur and moles of oxygen, so moles of sulphur, that is the mass of sulphur 60.0 gram divided by the molar mass of sulphur, which is 32.066 gram per mole, and we get 1.87 moles, then we'll find Out moles of oxygen, and that will be the mass of oxygen 40.0 gram, divided by the molar mass of oxygen, which is 16.0 gram mole. We cancel gram and gram and we get 2.5 moles now we'll find out the molar ratio of sulphur to oxygen. That is 1.8722 .5. Now we divide both numbers by the lowest number 1.87 and we get 121.33 now to make it a whole number is multiply 1 by 3 also 1.33 by 3. So we get the whole number ratio of sulphur to oxygen 3 to 4 point. So the empirical formula of this compound will be s 3, o 4, and that is the empirical formula.
Answer
Verified
Hint :In general, an empirical formula is the formula which gives the simplest whole number ratio of atoms of the various elements present in the molecule of the respective compound. The molecular formula divided by 2 gives an empirical formula.
Complete Step By Step Answer:
The steps to write empirical formula;
If we are given the mass percent of the elements present in a compound, then to find its empirical formula is easy.
Step 1. Conversion of mass percent to
grams.
Since we are having mass percent. It is convenient to use 100g of the compound as the starting material thus in a 100g sample of the above compound, 60g of sulphur and 40g of oxygen is present.
Step 2: Now we will convert into moles.
Moles of sulphur: $ \dfrac{{60g}}{{32.065}} = 1.87 $
Moles of oxygen: $ \dfrac{{40g}}{{16g}} = 2.5 $
Step 3: Divide the mole value, by smallest number
For sulphur, $ \dfrac{{1.87}}{{1.87}} = 1 $
For oxygen $ \dfrac{{2.5}}{{1.87}} =
1.336 \simeq 2 $
So, the compound is $ S{O_2} $ .
Additional Information:
Sulphur exists in sulphate form too. sulphate is widely used in industry, for example gypsum is a natural mineral which is a form of hydrated calcium sulphate. It is used to produce plaster. Copper sulphate, is a common algaecide, and is used as an electrolyte in galvanic cells. They are also used in therapeutic baths. They occur in nature as sulphate reducing bacteria, some anaerobic microbes, those
who live in deep sea thermal vents they reduce sulphate for chemosynthesis.
Note :
Sulphates are discharged into water from mines and smelters and from kraft pulp and paper mills, and many more industries. The atmospheric sulphur dioxides are formed by the combustion of fossil fuels, after that they are metallurgically roasted, which contributes to sulphate discharges into surface water.