
Empirical formula of a compound containing 60% sulphur and 40% oxygen?
Answer
506.4k+ views
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.
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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

