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EXAMPLE OF SOLUTION

1.To determine a relative molecular mass of Sodium Sulfate Na2SO4.

Relative molecular mass Ìr is determined as sum of relative atomic masses Àr of elements in this compound:

Mr(Na2SO4)=2Ar(Na)+Ar(S)+4Ar(O)=2×23+32+4×16=142 c.u.

2. To determine molar mass of Calcium Nitrite Ñà(NO3)2.

Molar mass of substance, expressed in grams, is numerically equal to relative molecular mass expressed in a.m.u. So, calculating relative molecular mass of Ñà(NO3)2, we obtain:

Mr(Ca(NO3)2)=Ar(Ca)+2Ar(N)+6Ar(O)=40+2×14+6×16=164 c.u.

And molar mass of Calcium Nitrite is equal to 164 g/mol.

3. How many atoms and moles are contained in 60 g of Carbon 12Ñ? Determine mass of one Carbon atom in grams.

If mass of substance m and its molar mass Mm are known, we could determine the amount of substance (number of moles) in this mass:

Mm is equal to 1 mol;

m –“- x mol; and ν = x = .

Since m=60 g, Ìm (Ñ)=12 g/mol, then ν =

Multiplying number of moles (5 moles) by Avogadro’s number NA we determine the number of atoms in given mass of substance:

 
 

N=ν×NA=5×6,02×1023=3,01×1024 atoms.

Next step - to be divisible molar mass of Carbon Ìm (Ñ)=12 g/mol by Avogadro’s number NA for calculation of mass of Carbon one atom in grams:

4. How many moles of Hydrobromide HBr are contained in 0,162 grams of this compound?

Determine a relative molecular mass of HBr:

Mr(HBr)=Ar(H)+Ar(Br)=1+80=81.

 
 

So, molar mass of HBr is equal to 81 g/mol. Calculate the number of moles in 0,162 g of HBr:

5. For neutralization of acid it was used 6 moles of Calcium Hydroxide Ca(OH)2. How many grams of Ca(OH)2 are equal to such number of moles?

Calculate a molar mass of Ca(OH)2:

Mr(Ca(OH)2)=40+2×16+2×1=74 g/mol.

Mass of 6 moles: m=υ×Mr =6 g×74 g/mol =444 g.

6. To calculate a mass part of Sodium in Sodium Sulfate Na2SO4.

Calculate a molar mass of Sodium Sulfate Ìr (Na2SO4) = 142 g/mol. 1 mole of Sodium Sulfate contains 2 moles of Sodium atoms. Its mass is equal to: m =2 Ar (Na) = 2×23 = 46 g.

Mass part w (Na) is the ratio of Sodium mass to total mass of substance:

7. To determine a mass part of waterless salt in crystalline hydrate.

 
 

Molar mass of crystalline hydrate of blue vitriol Mr(CuSO4×5H2O)= Mr(CuSO4)+5Mr(H2O)=160+90=250 g/mol? One mole of hydrate (250 g) contains 160 g of waterless salt. We calculate Mass part w (CuSO4) as ratio of Copper Sulfate mass to mass of hydrate:


Date: 2015-01-12; view: 962


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