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Norma-Jean [14]
3 years ago
8

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives the mass percent of 31

.57% and 5.30%. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of –5.30°C is recorded for a solution made by dissolving 10.8 g of the compound in 25.0 g water. Determine the molar mass and molecular formula of the compound. Assume that the compound is a nonelectrolyte.
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
8 0

Answer:

C4H8O6 is the molecular formula

The molar mass is 152 g/mol

Explanation:

Step 1: Data given

Mass % C = 31.57%

Mass % H = 5.30 %

Mass % O = 100 - 31.57 - 5.30 = 63.13 %

The freezing point = –5.30°C

Mass of the compound = 10.8 grams

Mass of water = 25.0 grams

compound is a nonelectrolyte.

Step 2: Calculate moles of elements

moles C = 31.57 grams/ 12.01 g/mol = 2.629 moles

moles H = 5.30 grams/ 1.01 g/mol = 5.25 moles

moles O = 63.13 grams / 16.0 g/mol = 3.95  moles

Step 3: Calculate the mol ratio

We divide by the smallest number

C: 2.629 moles / 2.629 moles = 1

H: 5.25 moles / 2.629 moles = 2

O: 3.95 moles / 2.629 moles = 1.5

The empirical formula is C2H4O3

The molecular mass of this is 76.06 g/mol

Step 4: Calculate moles solute

5.30 = m * 1.86

m = 2.85 = moles solute / 0.025 Kg

moles solute = 2.85 * 0.025 = 0.07125 moles

Step 5: Calculate molar mass

molar mass = 10.8 grams / 0.07125 moles = 152 g/mol

Step 6: Calculate molecular formula

152/76.06 ≈ 2

We have to multiply thr empirical formula by 2

2*(C2H4O3) = C4H8O6

C4H8O6 is the molecular formula

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The correct answer   for Mass of Na₂HPO₄ = 4.457 g and mass of  NaH₂PO₄  = 8.23 g

Given :  pH = 6.86

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Asked : Mass of  Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of  Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from  NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

<u>Step 1 : To find pka </u>

H₂PO₄⁻  <=> HPO₄²⁻  

The above reaction has pka = 7.2 ( from image shown )

<u>Step 2 : Plug values in Hasselbalch- Henderson equation </u>.

Hasselbalch -Henderson equation is to find pH  for buffer solution which is as follows :

pH = pka + log\frac{[A^-]}{[HA]}

pH = 6.86         pKa = 7.2

6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Subtracting  both side by 7.2

6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

-0.34 =  log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

Removing log

10^-^0^.^3^4 =   \frac{[HPO_4^2^-]}{[H_2PO_4^-]}

\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457 ---------------- equation (1)

<u>Step 3 : To find  molarity of H₂PO₄⁻ and HPO₄²⁻</u>

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence,  [H₂PO₄⁻ ] + [ HPO₄²⁻ ] =  0.1 M

Assume [H₂PO₄⁻ ] = x

So ,  [x ] + [ HPO₄²⁻ ] =  0.1 M

[ HPO₄²⁻ ] =  0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ]

[H₂PO₄⁻ ]  = x

 [ HPO₄²⁻ ] = 0.1 - x

Equation (1) = >\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457

Plug value of [H₂PO₄⁻ ]  and  [ HPO₄²⁻ ] ( from step 3 ) into equation (1)  as :

\frac{[0.1 - x ]}{[ x]} = 0.457

Cross multiplying

0.1 - x  = 0.457 x

Adding x on both side

0.1 -x + x = 0.457 x + x

0.1  = 1.457 x

Dividing both side by 1.457

\frac{0.1}{1.457} = \frac{1.457 x }{1.457}

x = 0.0686 M

Hence , [H₂PO₄⁻ ]  = x  = 0.0686 M

 [ HPO₄²⁻ ] = 0.1 - x

 [ HPO₄²⁻ ]  =   0.1 - 0.0686  

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of  H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M  or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M  or 0.0314 mole per 1 L

Since  that volume of buffer solution  is 1 L , so Molarity  = mole

Hence Mole of NaH₂PO₄  = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

<u>Step 6 : To find mass  of Na₂HPO₄  and NaH₂PO₄ </u>

Moles of  Na₂HPO₄  and NaH₂PO₄  can be converted to their masses using molar mass as follows :

Molar mass of  Na₂HPO₄  = 141.96 \frac{g}{mol}

Molar mass of NaH₂PO₄ = 119.98 \frac{g}{mol}

Mass (g) = mole (mol)* molar mass(\frac{g}{mol})

Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}

Mass of Na₂HPO₄ = 4.457 g

Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}

Mass of  NaH₂PO₄  = 8.23 g

5 0
3 years ago
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