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2 years ago

Empirical formula of a compound composed of 32.1 g potassium (k) and 6.57 g oxygen (o)?

Chemistry

1 answer:

tester [92]2 years ago

5 0

The empirical formula is K₂O.

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of K to O.

Step 1. Calculate the moles of each element

Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K

Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0

Step 2. Calculate the molar ratio of each element

Divide each number by the smallest number of moles and round off to an integer

K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1

Step 3: Write the empirical formula

EF = K₂O

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2 years ago

Brass is a substitutional alloy consisting of a solution of copper and zinc. A particular sample of red brass consisting of 79.0

Bas_tet [7]

Answer: The molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

Explanation:

Solute is the substance which is present in smaller proportion in a mixture and solvent is the substance which is present in larger proportion in a mixture.

We are given:

(m/m) % of Cu = 79 %

This means that 79 g of copper is present in 100 grams of alloy.

(m/m) % of Zn = 21 %

This means that 21 g of zinc is present in 100 grams of alloy.

As, zinc is present in smaller proportion. So, it is solute and copper is the solvent.

Calculating molality of zinc:

To calculate molality of the zinc, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$m_{solute}$ = Given mass of solute (Zinc ) = 21 g

$M_{solute}$ = Molar mass of solute (Zinc) = 65.3 g/mol

$W_{solvent}$ = Mass of solvent (copper) = 79 g

Putting values in above equation, we get:

$\text{Molality of Zinc}=\frac{21\times 1000}{65.3\times 79}\\\\\text{Molality of zinc}=4.07m$

Calculating molarity of zinc:

To calculate volume of a substance, we use the equation:

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

Density of solution = $8740kg/m^3=\frac{8740kg\frac{1000g}{1kg}}{1m^3\times \frac{10^6cm^3}{1m^3}}=\frac{8740g}{1000cm^3}=8.740g/cm^3$

(Conversion factors used are: 1 kg = 1000 g & $1m^3=10^6cm^3$ )

Mass of solution = 100 g

Putting values in above equation, we get:

$8.740g/cm^3=\frac{100.0g}{\text{Volume of zinc}}\\\\\text{Volume of zinc}=11.44cm^3$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molar mass of zinc = 65.3 g/mol

Volume of solution = $11.44cm^3=11.44mL$ (Conversion factor: $1cm^3=1mL$ )

Mass of zinc = 21.0 g

Putting values in above equation, we get:

$\text{Molarity of zinc}=\frac{21\times 1000}{65.3\times 11.44}=28.11M$

Hence, the molality and molarity of zinc in the solution is 4.07 m and 28.11 M respectively.

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2 years ago

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nlexa [21]

Answer:

Explanation:

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