All categories

3 years ago

Find the molecular weight of CO

Chemistry

2 answers:

boyakko [2]3 years ago

8 0

The molecular mass of CO is 12.01 + 16.00 or 28.01 g/mol.

Airida [17]3 years ago

6 0

Answer:

The molecular weight of CO is 28.01g/mole.

You might be interested in

A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

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 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

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 = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

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

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago

If you had .0451 moles of CaCl2, how many grams would you have?

Montano1993 [528]

Answer:

option C

Explanation:

because i took the test

8 0

3 years ago

HELP!!!!!! DUE IN 1 HOUR

Gnesinka [82]

Answer:

Group 17, 2, 7, 7 electrons

Explanation:

5 0

3 years ago

What causes the changes in air pressure on the earth surface

Fudgin [204]

Atmospheric Pressure

5 0

4 years ago

What is the molarity of a 17.0% by mass solution of sodium acetate, NaC2H3O2 (82.0 g/mol), in water? The density of the solution

sattari [20]

Answer:

[NaCH₃COO] = 2.26M

Explanation:

17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)

Let's determine the volume of solution, by density

Mass of solution / Volume of solution = Solution density

100 g / Volume of solution = 1.09 g/mL

100 g / 1.09 g/mL = 91.7 mL

17 grams of solute is contained in 91.7 mL

Molarity (M) = Mol of solute /L of solution

91.7 mL / 1000 = 0.0917L

17 g / 82 g/m = 0.207 moles

Molariy = 0.207 moles / 0.0917L → 2.26M

4 0

3 years ago