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A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL

AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

Answer: The molecular formula for the given organic compound is $C_6H_{14}O_2$

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and the freezing point of solution

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=i\times K_f\times m$

Or,

$\text{Freezing point of pure solution}-\text{freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution (water) = 0°C

Freezing point of solution = -3.21°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal boiling point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute = 10.20 g

$M_{solute}$ = Molar mass of solute = ?

$W_{solvent}$ = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

$(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g$

Calculating the molecular formula:

We are given:

Percentage of C = 60.98 %

Percentage of H = 11.94 %

Percentage of O = (100 - 60.98 - 11.94) % = 27.08 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 60.98 g

Mass of H = 11.94 g

Mass of O = 27.08 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{60.98g}{12g/mole}=5.082moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{11.94g}{1g/mole}=11.94moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{27.08g}{16g/mole}=1.69moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.69 moles.

For Carbon = $\frac{5.082}{1.69}=3$

For Hydrogen = $\frac{11.94}{1.69}=7.06\approx 7$

For Oxygen = $\frac{1.69}{1.69}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 7 : 1

The empirical formula for the given compound is $C_3H_7O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 118.2 g/mol

Mass of empirical formula = 59 g/mol

Putting values in above equation, we get:

$n=\frac{118.2g/mol}{59g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(7\times 2)}O_{(1\times 2)}=C_6H_{14}O_2$

Hence, the molecular formula for the given organic compound is $C_6H_{14}O_2$

8 0

3 years ago

Calculate the mass of solute in 500cm³ of 1.5mol/dm³ sodium hydroxide solution

sineoko [7]

Answer: The general formulae for moles is n=m/mr so now we have being given to find the mass so all we have to do is to change subject

Explanation: so we have to change subject in this question to m= n× mr . so in the question below we have being given the mole as 1.5mol/dm³ so all we have to do is to find the molecular relative mass(mr) .

to find the molecular relative mass of sodium hydroxide (NAOH) we add all of the atomic masses of all the atoms present so here we have sodium oxygen and hydrogen atoms present.

NA=23 O=16 H=1 so we add 23+16+1=40 so 40 is our molecular relative mass

now we fix it in our formulae which is m=n× mr

m=1.5× 40 =60 so our mass is 60grams or 60g

HOPE THIS HELPS!!!! if i made a mistake our MAY answer may be wrong feel free to comment

8 0

3 years ago

What statement does not accurately describe the following equation? C6H12O6+--->6H20+6CO2+energy A) oxygen is a react ant B)

STatiana [176]

The answer is B. The complete equation is C6H12O6 + 6O2 -->6H2O + 6CO2 + energy. So we can know that A and C and D is right. For B, the reaction release energy so it is exothermic reaction.

4 0

3 years ago

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The reaction of Pb+4 and O-2 produces the compound of

vovikov84 [41]

Answer:

Pb₂O₄

Explanation:

The given species are:

Pb⁴⁺ O²⁻

Now, to solve this problem, we use the combining powers which corresponds to the number of electrons usually lost or gained or shared by atoms during the course of a chemical combination.

Pb⁴⁺ O²⁻

Combining power 4 2

Exchange of valencies 2 4

Now the molecular formula is Pb₂O₄

4 0

2 years ago

When PCl5 solidifies it forms PCl4+ cations and PCl6– anions. According to valence bond theory, what hybrid orbitals are used by

Vlad1618 [11]

Answer:

sp³

Explanation:

Number of hybrid orbitals = ( V + S - C + A ) / 2

Where

H is the number of hybrid orbitals

V is the valence electrons of the central atom = 5

S is the number of single valency atoms = 4

C is the number of cations = 1

A is the number of anions = 0

For PCl₄⁺

Applying the values, we get:

H = ( 5+4-1+0) / 2

= 4

This corresponds to sp³ hybridization.

5 0

3 years ago

Read 2 more answers