Problem 121
Question
A chloride of a metal (M) has \(65.5 \%\) of chlorine. 100 \(\mathrm{ml}\) of vapour of the chloride of metal at STP weighs \(0.72 \mathrm{~g}\). The molecular formula of this metal chloride is (a) \(\mathrm{MCl}_{3}\) (b) \(\mathrm{MCl}_{4}\) (c) \(\mathrm{M}_{2} \mathrm{Cl}_{3}\) (d) \(\mathrm{MCl}_{3}\)
Step-by-Step Solution
Verified Answer
The molecular formula of the metal chloride is \(MCl_3\).
1Step 1: Determine Molar Mass of Chloride
From the given information, the chloride of metal contains 65.5% chlorine by mass. This implies that in 100g of the compound, 65.5g is chlorine, and 34.5g is metal M.
2Step 2: Calculate Molar Mass of the Compound
The mass of 100 ml vapor at STP can be used to determine the molar mass. At STP, 1 mol of gas is 22.4 L or 22400 ml. Using direct proportionality:\[\frac{0.72 \, ext{g}}{100 \, ext{ml}} = \frac{x \, ext{g}}{22400 \, ext{ml}} \]This results in:\[x = \frac{0.72 \, ext{g} \times 22400 \, ext{ml}}{100 \, ext{ml}} = 161.28 \, ext{g/mol}\]Hence, the molar mass of the chloride is 161.28 g/mol.
3Step 3: Calculate Molar Mass of Chlorine in Compound
The mass that is chlorine is 65.5% of 161.28 g/mol. Therefore:\[\text{mass of chlorine} = 0.655 \times 161.28 \, ext{g/mol} = 105.56 \, ext{g/mol}\]Now, calculate the amount of chlorine in moles:\[\frac{105.56}{35.5} \approx 3 \]Chlorine exists as \(Cl_3\) based on approximately 3 moles.
4Step 4: Determine Molar Mass of Metal M
Subtract the molar mass of chlorine from the total molar mass to find the molar mass of the metal M:\[\text{molar mass of M} = 161.28 \, ext{g/mol} - 105.56 \, ext{g/mol} = 55.72 \, ext{g/mol}\]Thus, the molar mass of metal M is approximately 55.72 g/mol.
5Step 5: Identify the Molecular Formula
We have \(MCl_3\) from our calculations since there are 3 moles of chlorine. Therefore, the molecular formula for the chloride of metal is \(MCl_3\).
Key Concepts
Molar Mass CalculationVapor DensityChlorine Percentage Composition
Molar Mass Calculation
Molar mass is a fundamental concept in chemistry used to determine how much one mole of a substance weighs. It is typically expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you need to know the atomic masses of the elements that make up the compound and how many atoms of each element are present.
When approaching a problem involving molar mass, follow these steps:
- Identify all the elements in the compound.
- Find the atomic mass of each element, usually available from the periodic table.
- Multiply the atomic mass of each element by the number of times the element appears in the compound.
- Add up all the masses to get the total molar mass of the compound.
Vapor Density
Vapor density is a useful parameter in determining the molar mass of gaseous substances. It is calculated as the mass of a certain volume of vapor compared to the same volume of hydrogen gas under the same conditions. However, it can also relate to how much a particular volume of vapor weighs in relation to standard conditions, usually at STP.
Understanding vapor density helps chemists infer the molar mass from a known vapor weight per unit volume. At STP, 1 mole of any gas occupies 22.4 liters. This standard allows the scientists to deduce relationships between the volume of gas measured and the molar mass.
In our problem, 100 ml of gas weighed 0.72 g. By scaling up this relationship to a full mole's volume (22,400 ml for gases at STP), we deduced the molar mass of the chloride compound. Essentially, multiplying the vapor density by the equation for STP volume allowed us to compute the mass of the entire mole of the chloride compound accurately.
Chlorine Percentage Composition
Percentage composition specifies how much of each element is in a compound by mass. For chlorine in this problem, this means calculating the fraction of the compound's mass that chlorine constitutes, converting it to a percentage. This is calculated by:
- Taking the mass of the element in 100g of the compound.
- Dividing it by the total molar mass of the compound.
- Multiplying by 100 to express it as a percentage.
Other exercises in this chapter
Problem 119
If \(3.02 \times 10^{19}\) molecules are removed from \(98 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), then the number of moles of \(\mathrm{H}_{2} \ma
View solution Problem 120
A metal oxide has the formula \(\mathrm{M}_{2} \mathrm{O}_{3} .\) It can be reduced by hydrogen to give free metal and water. \(0.1595 \mathrm{~g}\) of the meta
View solution Problem 122
The strength of \(0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) solution in terms of molality is nearly \(\quad\) when the density of the solution is \(1.1
View solution Problem 123
One mole of magnesium in the vapour state absorbed \(1200 \mathrm{~kJ} \mathrm{~mol}^{-1}\) of energy. If the first and second ionization energies of \(\mathrm{
View solution