Question

 

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Calculate the mass percent of cadmium in ,CdS cadmium(II) sulfide.

Mass percent = %

If you wish to obtain 14.0 g of cadmium metal from cadmium(II) sulfide, what mass of CdS (in grams) must you use?

Mass = g

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Nornicotine, a poisonous compound found in tobacco leaves, is 72.9% C , 8.16% H , and 18.94% N . Its molar mass is 148 g/mol. What are the empirical and molecular formulas of nornicotine?

Empirical formula:

Molecular formula:

Which of the following samples has the largest number of ions?

1.0 g of CaBr2
1.0 g of SrS
1.0 g of MgBr2
1.0 g of BeSO4
1.0 g of BaCO3

Answer 1

1. Molar mass of CdS =144.5 g/mol

1 mol of CdS contain 1 mol of Cd

Mass of 1 mol of Cd = 112.5g

Thus, mass percent of cadmium in ,CdS cadmium(II) sulfide.= 112.5/ 144.5 * 100 % = 77.8 %

Now, 1 mol of CdS contain 1 mol of Cd

i.e. 144.5 g of CdS contain 112.5 g of Cd

or 112.5 g of Cd can be obtained from 144.5 g of CdS

SO, 14 g of Cd can be obtained from (144.5 g * 14 g) / 112.5 g = 17.98 g of CdS

2. Let total mass of compound be 100 g

Thus, Mass of carbon in compond = 72.9 % of 100g = 72.9*100g/ 100 = 72.9 g

Molar mass of Carbon = 12 g/mol

Thus moles of Carbon in compound = 72.9g/(12g/mol) = 6.075 mol

Mass of Hydrogen in compond = 8.16 % of 100g = 8.16*100g/ 100 = 8.16 g

Molar mass of hydrogen = 1 g/mol

Thus moles of Hydrogen in compound = 8.16g/ ( 1g/mol)= 8.16 mol

Mass of nitrogen in compond = 18.94 % of 100g = 18.94*100g/ 100 = 18.94g

Molar mass of Nitrogen = 14 g/mol

Thus moles of Nitrogen in compound = 18.94 g /( 14g/mol) = 1.352 mol

Now, mole ration of Carbon, Nitrogen and hydrogen is

6.075 mol of C : 1.352 mol of nitrogen : 8.16 mol of hydrogen

Converting it into smple mole ratio

       

4.5 : 1: 6

9: 2: 12

Thus, EMpirical formula of Nornicotine is C₉H₁₂N₂

Molar mass of compound given = 148 g/mol

Mass of emipical formula of Nornicotine = (9* molar mass of C ) + ( 12* molar mass of Hydrogen ) +( 2* molar mass of Nitrogen) = (9* 12) + (12*1) +( 2* 14) = 148 g/mol

Here molar mass of compound = Mass of emipical formula of Nornicotine = 148g/ mol

Thus,

Empirical formula = Molecular formula = C₉H₁₂N₂

3. (a) Mass of CaBr₂ = 1.0 g

Molar mass of CaBr₂= 200g/mol

Thus, Moles of CaBr₂ =mass / molar mass = 1.0 g/ (200g/mol) = 0.005 mol

Number of molecules of CaBr₂ = moles * avagadro's number = 0.005 mol * 6.022* 10²³ molecules/ mol

= 3.011 * 10²¹ molecules

1 molecule of CaBr₂ gives 3 ions

Thus, total number of ions produced by 1.0 g of CaBr₂ is 3*  3.011 * 10²¹ molecules =  9.011 * 10²¹ ions

b) Mass of SrS = 1.0 g

Molar mass of  SrS = 119.7 g/ mol

Thus, Moles of SrS =mass / molar mass = 1.0 g/ (119.7g/mol) = 0.008 mol

Number of molecules of SrS = moles * avagadro's number = 0.008 mol * 6.022* 10²³ molecules/ mol

= 5.031 * 10²¹ molecules

1 molecule of  SrS gives 2 ions

Thus, total number of ions produced by 1.0 g of SrS is 2* 5.031 * 10²¹ molecules = 1.006 * 10²² ions

(c) Mass of MgBr2 = 1.0 g

Molar mass of MgBr2= 184g/mol

Thus, Moles of MgBr2 =mass / molar mass = 1.0 g/ (184g/mol) = 0.0054 mol

Number of molecules of MgBr2 = moles * avagadro's number = 0.0054 mol * 6.022* 10²³ molecules/ mol

= 3.273 * 10²¹ molecules

1 molecule of MgBr2 gives 3 ions

Thus, total number of ions produced by 1.0 g of MgBr₂ is 3*  3.273 * 10²¹ molecules =  9.818 * 10²¹ ions

d) Mass of BeSO4= 1.0 g

Molar mass of BeSO4 = 105 g/ mol

Thus, Moles of BeSO4 =mass / molar mass = 1.0 g/ (105g/mol) = 0.0095 mol

Number of molecules of BeSO4 = moles * avagadro's number = 0.0095 mol * 6.022* 10²³ molecules/ mol

= 5.735 * 10²¹ molecules

1 molecule of BeSO4 gives 2 ions

Thus, total number of ions produced by 1.0 g of BeSO4 is 2* 5.735 * 10²¹ molecules = 1.147 * 10²² ions

e) Mass of BaCO3=  1.0 g

Molar mass of = 197 g/ mol

Thus, Moles of BaCO3 =mass / molar mass = 1.0 g/ (197g/mol) = 0.0051 mol

Number of molecules of BaCO3 = moles * avagadro's number = 0.0051 mol * 6.022* 10²³ molecules/ mol

= 3.057 * 10²¹ molecules

1 molecule of BaCO3 gives 2 ions

Thus, total number of ions produced by 1.0 g of BaCO3 is 2* 3.057 * 10²¹ molecules = 6.114 * 10²² ions

Thus, 1.0 g of MgBr2 has the largest number of ions