What is the molar mass of ethanol C2H5OH?

As the chemical formula for methanol or methyl alcohol is CH3OH. Molar mass of compound methanol (CH3OH)= sum of atomic masses of elements of the compound methanol (CH3OH).
So molar mass of methanol=12 + (1×3) + 16 + 1
=32g

Long answer:

Ethanol, C2H5OH, has 2 carbon atoms, 6 hydrogen atoms, and one oxygen atom.

Firstly, what is molar mass? Molar mass is the amount of grams in one mole of substance. A mole is a quantity of substance. It was defined so that every mole of carbon weighs 12 grams. Which gives us $6,022x10²²$ atoms of carbon. And if we take $6,022x10²²$atoms of hydrogen, it would give us 1 gram, and 16 grams for every mole of oxygen.

If there’s 2 carbon atoms per molecule of ethanol, we do the math:

12 grams per mole x 2 atoms per molecule = 24 grams of carbon per mole of molecules

Same for hydrogen and oxygen:

1 gram per mole x 6 atoms per molecule = 6 grams of hydrogen per mole of molecules

16 grams per mole x 1 atom per molecule = 16 grams of oxygen per mole of molecules

Now we add 24 grams of carbon + 6 grams of hydrogen + 16 grams of oxygen per mole , it gives us a total of 46 grams per mole of ethanol.

If every mole of ethanol has 46 grams, then that is the molar mass of ethanol.

Short answer:

Ethanol has a molar mass of 46 grams.

The molar mass (M) of a substance is the mass of one mole of that substance. For a molecule, the molar mass is calculated by multiplying the subscript of each element by its molar mass, which is its atomic weight on the periodic table in g/mol.

C2H5OH is the formula for ethanol. The molar masses of each element in the ethanol molecule are:

M C = 12.011 g/mol

M H = 1.008 g/mol

M O = 15.999 g/mol

When calculating the molar mass of an alcohol, the subscripts for hydrogen can be combined. The formula for ethanol can also be written as C2H6O.

M C2H5OH = (2 × 12.011 g/mol) + (6 × 1.008 g/mol) + (1 × 15.999 g/mol) = 46.069 g/mol

Answer: The molar mass of ethanol, C2H5OH, is 46.069 g/mol.

Okay so am assuming that air and C2H2 are at the same pressure and temperature.

Also %of oxygen in air as 21%

According to avagadro’s law,

At same P and T equal volumes of gases will contain equal number of moles.

Let the number of moles in 8litres of C2H2 be “x”

Burning of C2H2 goes like this

C2H2 + (5/2)O2 = 2CO2 + H2O

So for 1mole of C2H2 you need 5/2 moles of O2

So 8lites of C2H2 requires 8×5/2=20 lites of O2

Now only 21% of air is oxygen so you will need,

(20/21)× 100 litres of air to get 20 litres of oxygen.

So 95.24L of air is required for burning of 8L of C2H2.

Mass = density × volume

Volume of ethanol = 1.0 L = 1000 mL

Density = 0.789 g/mL

Mass of ethanol = 0.789 g/mL × 1000 mL

= 789 g

Molar mass of ethanol (C2H5OH) = 46 g/mol

No. of moles in 789 g of ethanol = (789/46) mol

= 17.15 mol (answer)

C2H5OH is the chemical formula for ethanol.

Molar mass of C2H5OH = 46.07 g/mol

Mass of C2H5OH = 10g.

Moles = mass/molar mass = 10g/46.07g/mol

Moles =0.217mol

Formula of ethanol is C2H5OH.

2(C) = 2(12) = 24

5(H) = 5(1) = 5

1(O) =1(16) = 16

1(H) =1(1)= 1

Add these , you will get 46g/mol ( ✓answer )

( NOTE : 12 - mass no of Carbon /

1 - mass no of Hydrogen /

16 - mass no of Oxygen )

Thanks for your question…

25 percent solution by mass of ethanol means that 100 gram of solution contains 25 gram of solute i.e-ethanol.Now,as the density of the solution is 1 g/ml so 100 g of the solution has a volume of 100 ml.

So,we have 25 gram of ethanol in 100 ml of the solution. So we have 250 g of ethanol in 1 litre of solution.Now molar mass of ethanol is 46. So,250 g of ethanol is equal to 250/46 moles of ethanol which is approximately 5.43 moles of ethanol.So, we have that 1 litre of the given solution contains 5.43 moles of ethanol.

So the molarity of the given solution is 5.43

You need to know the formula of sodium nitrate. Sodium ion has a plus 1 charge and therefore combines in a 1:1 ratio with the nitrate ion which is -1.

So the formula is NaNO3.

The molar mass is then the sum af the atomic masses of

1 Na + 1N + 3 O .