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Irina-Kira [14]

2 years ago

7

Usually potassium hydrogen phthalate is kept very pure. But Stu Dent thinks the bottle of potassium hydrogen phthalate has been

accidently mixed with sodium chloride salt. He decides to use a titration to determine the amount of potassium hydrogen phthalate present. He weighs 2.3854 g of the contaminated potassium hydrogen phthalate mixture. The titration uses 17.47 mL of 0.5000 M sodium hydroxide. Use the titration volume and concentration to calculate the moles of potassium hydrogen phthalate actually reacted. Then use the molar mass (204.2 g/mol) to calculate the mass of potassium hydrogen phthalate actually reacted. Enter the mass to 4 significant digits with units of g.

1 answer:

Ahat [919]2 years ago

5 0

Answer:

1.784 g

Explanation:

The equation of the reaction is;

NaOH(aq) + KHC8H4O4(aq) --------> KNaC8H4O4(aq) + H2O(l)

Number of moles of NaOH reacted = 17.47/1000 * 0.5000 M

Number of moles of NaOH reacted =8.735 * 10^-3 moles

From the reaction equation;

1 mole of NaOH reacted with 1 mole of KHC8H4O4

Hence, 8.735 * 10^-3 moles of NaOH reacts with 8.735 * 10^-3 moles of KHP.

So,

Mass of KHP reacted = 8.735 * 10^-3 moles * 204.2 g/mol = 1.784 g

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This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

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First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

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Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

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$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

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$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

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Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

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