If 2.54 g of zinc sulfide (Zn S) contains 1.70 g of Zn, what is the mass ratio o

Question

If 2.54 g of zinc sulfide (Zn S) contains 1.70 g of Zn, what is the mass ratio of zinc to sulfur? you must first determine the mass of sulfur in order to find the requested ratio ) How many kilograms each of element, Zn and S. are in 382 g of Zn S? the mass units are different What is the name of the fundamental law that you applied to solve part b? A chemist finds that 30.82 g of nitrogen will react with 17.60 g, 35.2 g, 70.40 g, or 88.00 g of oxygen to form four different compounds. Calculate the mass ratio of oxygen to nitrogen in each compound. Show whether the results from part a are consistent with the "Law of Multiple Proportions".

Explanation / Answer

1) a) 2.54 g of ZnS contains 1.70 g of Zn . So , mass of Sulphur (S) = 2.54-1.70 =0.84 g Hence , mass ratio = 1.70/0.84 = 2.02381 .

b) Molar weight of ZnS is 97.44 g/mol , which means 97.44 gm of ZnS conatins 65.38 g of Zn and 32.06 g of Sulphur . thoes are their respective atomic weights .

Hence , 382 g of ZnS contains 382 * 65.38 / 97.44 = 256.313g = 0.256313kg of Zn and 382 * 32.06 / 97.44 = 125.686 g = 0.125686 kg of S.

c) Avagadro's Hypothesis.

2) 30.82g of nitrogen(N) eacts with 17.60g , 35.20g , 70.40g , 88.00g of Oxygen(O) .

a) hence, tha mass ratios would simple be the (weight of Oxygen)/(weight of Nitrogen) in each case . The resulting values would be i) 0.571 , ii) 1.1421 , iii) 2.2842 , iv) 2.8553 .

b) Now we have to calculate the ratio of mass ratios obtained each case , which is given by 0.571 : 1.1421 : 2.2842 : 2.8553 . Now , dividing all the numbers by the least one (i.e., 0.571) , we get the ratio as 1 : 2 : 4 : 5 (approximately) .

These integer ratios suggest that the " law of Multiple proportions " is proved

  

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