Question 1 (1 point)

5.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 7

slega [8]

Answer:

The molecular formula = C6H6

Explanation:

Step 1: Data given

Mass of compound X = 5.00 grams

Mass of products =

CO2 = 16.39 grams

H2O = 3.46 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16. 0g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles CO2 = 16.93 grams / 44.01 g/mol

Moles CO2 = 0.385 moles

Moles C = 1* 0.385 = 0.385 moles

Moles H2O = 3.46 grams / 18.02

Moles H2O = 0.192 moles

Moles H = 2* 0.192 = 0.384 moles

Step 3: Calculate mass

Mass = moles * molar mass

Mass C = 0.385 moles *12.0 g/mol

Mass C = 4.62 grams

Mass H = 0.39 grams

Mass O = 5.00 - 4.62 -0.38 moles

Mass O = 0 grams

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.385 moles / 0.384 = 1

H: 0.384 moles / 0.384 = 1

The empirical formula is CH

This molecular formula is 13 g/mol

We have to multiply the empirical formula by n

n = 78 g/mol / 13 g/mol

n = 6

The molecular formula = 6*(CH) = C6H6

4 0

3 years ago

A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

<u>Explanation:</u>

To calculate the molarity of solute, 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:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$