1. The percentage of a composition is equal to ???.
2. A compound's molecular formaul is numerically equal to its ???
Help!! Thanks =]
Your questions lack specificity. They are not complete.
percent composition of what??
#2 is used in conjunction with what.
sorry i meant to type:
the percentage composition of a compound is equal to __________.
for # 2 the whole question is
a.) A compound's molecular formula is numerically equal to its _______
b.) THis means that the molecular formula can be determined given a compound's ___formula mass___
1. The percentage of a composition is equal to the ratio of the amount of a specific component to the total amount of the composition, multiplied by 100. In other words, it represents the proportion of a particular substance in a mixture or compound.
To calculate the percentage composition of a substance, you need to know the masses of the individual components. Here's the formula:
Percentage Composition = (Mass of Component / Total Mass of Composition) * 100
For example, if a compound has a total mass of 100 grams and one of its components has a mass of 40 grams, the percentage composition of that component would be:
(40 g / 100 g) * 100 = 40%
2. A compound's molecular formula is numerically equal to its empirical formula multiplied by the ratio of molar mass to empirical formula mass.
To find the empirical formula of a compound, you need to determine the simplest whole number ratio of atoms in the molecule. This can be achieved by dividing the number of atoms of each element by the greatest common divisor.
Once you have the empirical formula, you can find the molecular formula by multiplying it by a whole number, which represents the ratio of the molar mass to the empirical formula mass. This can be calculated by dividing the molar mass of the compound by the empirical formula mass.
For example, if a compound's empirical formula is CH2O and its molar mass is 180 g/mol, you can calculate its molecular formula as follows:
Empirical Formula Mass:
Carbon (C): 12.01 g/mol * 1 = 12.01 g/mol
Hydrogen (H): 1.01 g/mol * 2 = 2.02 g/mol
Oxygen (O): 16.00 g/mol * 1 = 16.00 g/mol
Empirical Formula Mass = 30.03 g/mol
Ratio of Molar Mass to Empirical Formula Mass:
Molar Mass: 180 g/mol
Empirical Formula Mass: 30.03 g/mol
Ratio = 180 g/mol / 30.03 g/mol = 5.99 ≈ 6
Therefore, the compound's molecular formula would be (CH2O) * 6 = C6H12O6.