A given sample of a xenon fluoride compound contains molecules of the type XeFn where n is some whole number. Given that 9.03 × 1020 molecules of XeFn weighs 0.368 g, determine the value for n in the formula.
To determine the value for n in the formula XeFn, we can use the given information that 9.03 × 1020 molecules of XeFn weighs 0.368 g.
1. Convert the given number of molecules to moles:
- We know that 1 mole of any compound contains Avogadro's number of molecules, which is approximately equal to 6.022 × 10^23.
- Therefore, the number of moles of XeFn can be calculated as follows:
n moles = (9.03 × 10^20 molecules) / (6.022 × 10^23 molecules/mole)
= 1.50 × 10^-3 moles
2. Use the molar mass of XeFn to determine the molar mass of one molecule:
- The molar mass of xenon (Xe) is approximately 131.293 g/mol.
- The molar mass of fluorine (F) is approximately 18.9984 g/mol.
- Since there are n fluorine atoms in one molecule of XeFn, the molar mass of XeFn is calculated as follows:
Molar Mass = (1 × molar mass of Xe) + (n × molar mass of F)
= 131.293 + (n × 18.9984) g/mol
3. Use the molar mass and the number of moles to calculate the weight of XeFn:
- The weight of XeFn can be calculated using the formula:
Weight = Molar Mass × Moles
= (131.293 + (n × 18.9984)) × 1.50 × 10^-3 g
4. Set up an equation based on the given information:
- According to the problem, 9.03 × 10^20 molecules of XeFn weigh 0.368 g, so we can write the equation as:
Weight = 0.368 g
- Substituting the calculated weight from step 3 into the equation:
(131.293 + (n × 18.9984)) × 1.50 × 10^-3 g = 0.368 g
5. Solve the equation for n:
- Distribute 1.50 × 10^-3 g to both terms:
196.94 + n × 27.9976 = 0.368 / (1.50 × 10^-3)
196.94 + n × 27.9976 = 245.3333333
- Subtract 196.94 from both sides:
n × 27.9976 = 48.3933333
- Divide both sides by 27.9976:
n ≈ 1.728
Therefore, the value of n in the formula XeFn is approximately 1.728.
To determine the value of n in the formula XeFn, we need to use the given information that 9.03 × 1020 molecules of XeFn weigh 0.368 g.
To solve this problem, we can use the concept of molar mass. Molar mass is the mass of one mole of a substance, expressed in grams per mole. The molecular weight of a compound can be calculated by adding up the atomic weights of all the atoms in the molecule.
1. Calculate the molar mass of XeFn:
- Xenon (Xe) has an atomic weight of 131.29 g/mol.
- Fluorine (F) has an atomic weight of 18.99 g/mol.
- Since there are n fluorine atoms in each molecules, the molar mass of XeFn is given by: (131.29 + 18.99n) g/mol.
2. Convert the given mass of 0.368 g to moles:
- To convert from mass to moles, we divide the given mass by the molar mass:
Moles = Mass (g) / Molar mass (g/mol)
- In this case, Moles = 0.368 g / (131.29 + 18.99n) g/mol.
3. Use Avogadro's number to convert from moles to molecules:
- Avogadro's number (6.022 × 1023) represents the number of molecules or atoms in one mole of substance.
- In this case, the number of molecules (N) is given by:
N = Moles × Avogadro's number
- Substituting the value of Moles calculated in step 2, we get:
N = (0.368 / (131.29 + 18.99n)) × (6.022 × 1023)
4. Since we are given that there are 9.03 × 1020 molecules of XeFn, we can equate N to this value and solve for n:
- (0.368 / (131.29 + 18.99n)) × (6.022 × 1023) = 9.03 × 1020
- Simplifying the equation, we get:
(0.368 × 6.022 × 1023) / (131.29 + 18.99n) = 9.03 × 1020
- Cross-multiplying and rearranging, we obtain:
(0.368 × 6.022 × 1023) = 9.03 × 1020 × (131.29 + 18.99n)
(0.368 × 6.022 × 1023) = (9.03 × 1020) × (131.29 + 18.99n)
(0.368 × 6.022 × 1023) = (9.03 × 1020) × 131.29 + (9.03 × 1020) × 18.99n
(0.368 × 6.022 × 1023) - (9.03 × 1020) × 131.29 = (9.03 × 1020) × 18.99n
(0.368 × 6.022 × 1023) - (9.03 × 1020) × 131.29 = (9.03 × 1020) × 18.99n
[(0.368 × 6.022 × 1023) - (9.03 × 1020) × 131.29] / [(9.03 × 1020) × 18.99] = n
By evaluating the above expression, we can determine the value of n in the formula XeFn.
Xe = 131 g/mol
F = 19 g/mol
so XeFn = (131 + 19 n) g/mol
or
(131+19n)g / 6*10^23 molecules
but this stuff is
.368 g/9.03*10^20 molecules
.368/9.03*10^20 = (131+19n) /6*10^23
131+19 n = 244.5 g/mol
19 n = 113.5
n = 6
so
XeF6