A student made a determination of the empirical formula ot tungsten oxide (WO) produced by the total ignition of air of a smaple of metallic Tungsten. From the lab:
Weight of crucible 11.120g
weight of tungsten 8.820
weight of crucible and product (WO) 22.836
a. Calculate the percent comp by mass of tengsten oxide
b. Determine the empirical formula of tungsten oxide.
Really having trouble with this. Product weighs 11.719 is all I can get started.
I just worked the problem and got 79% for W and 21% for O.
For the empirical, I got WO4.
Is this correct?
If you used 79% and 21% you should have obtained WO3, I think, however, I have
(8.82/11.716)*100 = 75.3%W and
(2.899/11.716)*100 = 24.7% O.
Using these numbers I get a ratio of 1W to 3.77O Oxygen and that makes no sense for the whole number ratio would be W4O15.
There is a W2O3, a WO2, and a WO3.
To solve this problem, we need to follow a step-by-step approach. Let's begin:
Step 1: Calculate the mass of oxygen in the tungsten oxide product.
To find the mass of oxygen, we need to subtract the mass of tungsten from the total mass of the tungsten oxide product.
Mass of tungsten = weight of crucible + weight of tungsten - weight of crucible and product = 11.120 g + 8.820 g - 22.836 g = -2.896 g (Note: The negative value indicates an error in the calculation or data entry. Please recheck and provide accurate values if necessary.)
Step 2: Calculate the percent composition by mass of tungsten oxide.
To calculate the percent composition by mass, we need to determine the mass of tungsten and oxygen separately and then divide them by the total mass of the tungsten oxide product.
Percentage of tungsten (W) = (mass of tungsten / mass of tungsten oxide product) × 100%
Percentage of oxygen (O) = (mass of oxygen / mass of tungsten oxide product) × 100%
However, as step 1 yielded a negative mass of tungsten, we cannot proceed with this calculation. Please recheck and provide accurate values for the weights involved.
Step 3: Determine the empirical formula of tungsten oxide.
The empirical formula represents the simplest whole number ratio of elements in a compound. To determine the empirical formula, we need the molar masses of tungsten (W) and oxygen (O).
The molar mass of tungsten (W) is 183.8 g/mol.
The molar mass of oxygen (O) is 16.00 g/mol.
Since we do not have accurate values for the mass of tungsten and oxygen, it is not possible to determine the empirical formula.
Please ensure you have accurate measurements for the weights involved, and I would be happy to provide further assistance.
To solve this problem, we need to use the concept of molecular weight and the Law of Conservation of Mass. Let's go step by step to answer each part of the question:
a. Calculate the percent composition by mass of tungsten oxide (WO):
To find the percent composition, we need to determine the mass of tungsten and oxygen in the compound.
1. Determine the mass of tungsten:
Weight of crucible = 11.120g
Weight of tungsten = 8.820g
Mass of tungsten = Weight of crucible and tungsten - Weight of crucible
= (8.820g + 11.120g) - 11.120g
= 8.820g
2. Determine the mass of oxygen:
Weight of crucible and product (WO) = 22.836g
Mass of oxygen = Weight of crucible and product (WO) - Mass of tungsten
= 22.836g - 8.820g
= 14.016g
3. Calculate the percent composition of tungsten oxide:
Percent Composition of Tungsten = (Mass of Tungsten / Mass of Tungsten Oxide) * 100%
= (8.820g / 22.836g) * 100%
= 38.6%
Percent Composition of Oxygen = (Mass of Oxygen / Mass of Tungsten Oxide) * 100%
= (14.016g / 22.836g) * 100%
= 61.4%
Therefore, the percent composition by mass of tungsten oxide is approximately 38.6% tungsten and 61.4% oxygen.
b. Determine the empirical formula of tungsten oxide:
Now, we need to find the empirical formula of tungsten oxide using the percent composition by mass.
1. Calculate the moles of each element:
Moles of Tungsten = (Mass of Tungsten / Atomic Weight of Tungsten)
= (8.820g / 183.84 g/mol) (Atomic weight of tungsten is 183.84 g/mol)
= 0.048 mol
Moles of Oxygen = (Mass of Oxygen / Atomic Weight of Oxygen)
= (14.016g / 16.00 g/mol) (Atomic weight of oxygen is 16.00 g/mol)
= 0.876 mol
2. Determine the ratio of moles:
Divide both moles by the lowest mole value:
Tungsten:Oxygen = 0.048 mol / 0.048 mol : 0.876 mol / 0.048 mol
= 1 : 18.25
3. Find the whole number ratio:
Multiply both sides by a factor to get whole numbers:
Tungsten:Oxygen = 1 : 18.25 ≈ 1 : 18
The empirical formula of tungsten oxide is WO18.
Note: Since WO18 is not a simplified ratio, we could divide both subscripts by the greatest common divisor (GCD). In this case, GCD(1, 18) = 1. Therefore, the simplified empirical formula is WO18.