If 9.03 x 1021 molecules of a compound having general formula CnH2n+2 weigh 1.71g, what is the molar mass of the compound? Give its molecular formula
mols = 9.03E21/6.02E23 = 0.015
mol = grams/molar mass
molar mass = grams/mol = 1.71/0.015 = about 114 for molar mass. .
CnH2n means there will be a CH3 on each end with those in the middle being 14. How many units of 14 do we have? That will be 114-30(two ends) = 84 and 84/14 = 6; therefore, the formula must be C8H18
To find the molar mass of the compound, we need to follow these steps:
Step 1: Calculate the number of moles
The number of moles can be found using the relationship between the number of molecules and Avogadro's constant (6.022 x 10^23 molecules per mole). We can calculate the number of moles using the formula:
Number of moles = Number of molecules / Avogadro's constant
Given that you have 9.03 x 10^21 molecules, we can substitute these values into the formula:
Number of moles = 9.03 x 10^21 / (6.022 x 10^23)
Step 2: Calculate the mass of the compound
The molar mass of a compound is the mass of one mole of that compound. The formula to calculate the mass is:
Mass = Number of moles x Molar mass
Given that you have 1.71g of the compound and you have calculated the number of moles, we can now solve for the molar mass:
1.71g = Number of moles x Molar mass
Rearranging this equation, we get:
Molar mass = 1.71g / Number of moles
Step 3: Calculate the molar mass of the compound
Now we can substitute the values into the formula to find the molar mass:
Molar mass = 1.71g / (9.03 x 10^21 / (6.022 x 10^23))
Simplifying this equation:
Molar mass = 1.71g x (6.022 x 10^23) / (9.03 x 10^21)
Molar mass ≈ 11.4 g/mol
Therefore, the molar mass of the compound is approximately 11.4 g/mol.
To determine the molecular formula of the compound, we need to know the values of n. The general formula CnH2n+2 suggests that each carbon atom is associated with two hydrogen atoms and that there are n carbon atoms. Without knowing the specific value of n, we cannot determine the exact molecular formula of the compound.