Answer to Question #160276 in General Chemistry for Smayya Ruganyira

Question

Calculate the mass percentage of water of crystallization in CuSO4.H2O. (Atomic mass: Cu = 64; S = 32 O= 16; H =1)  

 An alkane X of formula mass 3O consists of 80% carbon by mass.

a) Determine the empirical formula of X. 

b) Determine the molecular formula of X. 

c) write a chemical equation for the complete combustion of X in oxygen.

Ethane can react to form a solid whose molecular mass is more than 10,000. 

a) Name the reaction that occurs

b) Write the chemical equation of that reaction.

c) (i) State what is observed when ethane reacts with brornine.   

(ii) Write the equation for fhe reaction in 7(c)(i) above.

Explain why: a) Hard water requires a lot of soap to form a lather.

b) Isotopes of an element show similar chemical reactions.

c) When carbon dioxide is bubbled through lime water, the lime water turns milky and finally clears.

Calculate the mass of nitric acid (HNO3) required for preparing 200cm3of 2M HNO3 solution. (Atomic mass: H= 1, N= 14, O= 16).

When hydrogen gas was passed over x g of strongly heated copper (II) oxide until there was no further change, 4 g of a solid was formed. (Atomic mass: Cu = 64, O: 16)

a) State what was observed.

b) Write the equation for the reaction.

c) Determine the value of X.

Describe the industrial preparation of nitric acid from ammonia (the diagram is not required). Your description should include equations for reactions that occur.

The formation of methanol from hydrogen and carbon monoxide is represented by the equation:2H2(g) + CO(g)        CHsOH(g) ; the energy released = 92kJ /mol.Calculate the energy that is released, in kJ/mol, when 96g of methanol is formed. (Relative mass: C= 12, H= 1 and o= 16)  

Answer 1

Q160276

Calculate the mass percentage of water of crystallization in CuSO4.H2O. (Atomic mass: Cu = 64; S = 32 O= 16; H =1)  

Solution:

molar mass of CuSO₄.H₂O

= 1* atomic mass of Cu + 1* atomic mass of S + 5 * atomic mass of O + 2 * atomic mass of H.

= 1 * 63.546g/mol + 1*32.065g/mol + 5 *15.999 g/mol + 2 * 1.0079 g/mol

= 63.546g/mol + 32.065g/mol +79.995 g/mol +2.0158 g/mol

= 177.6218 g/mol

molar mass of H₂O = 2 * atomic mass of H + 1 * atomic mass of O

= 2 * 1.0079 g/mol + 1 * 15.999 g/mol

= 2.0158 g/mol + 15.999 g/mol

= 18.0148g/mol

Mass % of the water of crystallization = $=\frac{mass \space of \space H_2O}{mass \space of \space CuSO_4.H_2O}*100 = \frac{18.0148g/mol}{177.6218g/mol}*100$

$= \frac{1801.48}{177.6218} = 10.14$ %

Hence mass % of the water of crystallization of CuSO₄.H₂O is 10.14 %

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An alkane X of formula mass 3O consists of 80% carbon by mass.

a) Determine the empirical formula of X. 

b) Determine the molecular formula of X. 

c) write a chemical equation for the complete combustion of X in oxygen.

Solution:

The General chemical formula of alkanes is C_nH_2n+2 where n = 1, 2, 3,4, .......

Alkanes contain only Carbon and hydrogen.

So if the given compound contains 80% carbon, then the remaining 20% will be hydrogen.

Let us assume that the mass of the compound is 100 grams.

So there will be 80grams of Carbon and 20 grams of Hydrogen in the given compound.

a) Determine the empirical formula of X. 

Convert 80g of C and 20 g of H to moles.

moles of Carbon = $= 80 g\space of\space C * \frac{1\space mol \space C}{12.011g\space of \space C } = 6.66\space mol \space of \space C$

moles of Hydrogen = $= 20 g\space of\space H * \frac{1\space mol \space H}{1.0079g\space of \space H } = 19.84 \space mol \space of \space H$

Among the given mole of C and H, carbon is the one with the least number of moles, so divide moles of

both of them by moles of C.

For C $=\frac{6.66 mol \space of \space C}{6.66 mol \space of \space C} = 1$

For H $=\frac{19.84 mol \space of \space H}{6.66 mol \space of \space of C} = 2.979 = 3$

So the Empirical formula of the given Alkanes is C₁H₃. which can also be written as CH₃.

b) Determine the molecular formula of X. 

Solution: The molar mass of the compound given in the question = 30 g/mol

The empirical formula mass of CH₃ = 1 * atomic mass of C + 3 * atomic mass of H

= 1 * 12.011 g/mol + 3 * 1.0079 g/mol

= 12.011 g/mol + 3.0237g/mol

= 15.0347g/mol

The ratio of molar mass and empirical formula mass will give us integer i , which when multiplied to subscripts of empirical formula will give us the molecular formula of the compound.

i = $=\frac{molar\space mass }{Empirical \space formula\space mass } = \frac{30}{15.0347} = 2 ;$

So the molecular formula of the compound is = C_2*1 H_2*3 = C₂H₆.

c) write a chemical equation for the complete combustion of X in oxygen.

Solution:

C₂H₆ will react with O₂ to form CO₂ and H₂O.

C₂H₆ (g) + O₂ (g) ==> CO₂ (g) + H₂O(g)

We will have to balance this reaction.

There is 2 C on the left and 1 C on right.

Balance the C by writing 2 in front of CO2

C₂H₆ (g) + O₂ (g) ==> 2CO₂ (g) + H₂O(g)

Now there is 6 H on the left and 2 H on right. Balance the Hydrogen by writing 3 in front of H₂O.

C₂H₆ (g) + O₂ (g) ==> 2CO₂ (g) + 3H₂O(g)

Lastly, we balance the O. There is 2 O on the left and 7 O on right.

To balance the O we will have to write a proper coefficient in front of O2.

Here we will have to use the fraction 7/2 in front of O2 for balancing O on both the side.

C₂H₆ (g) + 7/2 O₂ (g) ==> 2CO₂ (g) + 3H₂O(g)

Convert the fraction 7/2 to a whole number by multiplying the whole reaction by 2.

2C₂H₆ (g) + 7 O₂ (g) ==> 4CO₂ (g) + 6H₂O(g)

This is our balanced chemical reaction for the combustion of ethane C₂H₆.

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Calculate the mass of nitric acid (HNO₃) required for preparing 200cm³ of 2M HNO₃ solution. (Atomic mass: H= 1, N= 14, O= 16).

Solution:

molar mass of HNO3 = 1 * atomic mass of H + 1 * atomic mass of N + 3 * atomic mass of O.

= 1 * 1.0079 g/mol + 1 * 14.007g/mol + 3 * 15.999 g/mol

= 1.0079g/mol + 14.007g/mol + 47.997g/mol

= 63.012 g/mol

In the question, we are given 200 cm³ of 2M HNO₃ solution.

The formula of molarity is

$Molarity = \frac{moles \space of \space HNO_3}{volume\space of \space HNO_3 \space in \space 'L' }$

Convert 200cm³ to 'L' by using the conversion factor 1cm³ = 1mL and 1L = 1000mL.

volume of HNO₃ in 'L' = 200cm³ * 1mL/ 1cm³ * 1L/1000mL = 0.200L.

plug 0.200L and 2M in the molarity formula we have

$2M = \frac{moles \space of \space HNO_3}{0.200L };$

After solving it we have

moles of HNO₃ = 0.400 moles .

Grams of HNO₃ = moles x molar mass = 0.400 mol * 63.012 g/mol

= 25.205 grams.

Hence we will require 25.2 grams of HNO₃ for preparing the given solution of HNO₃.