1) A compound has the following percentages by mass: barium, 58.84%; sulfur, 13.74%; oxygen, 27.43%. Determine the empirical formula of the compound.
2) If a 1.271-g sample of aluminum metal is heated on a chlorine gas atmosphere, the mass of aluminum chloride produced is 6.280 g. Calculate the empirical formula of aluminum chloride.
I don't understand these at all, could you at least help explain them/push me in the right direction on how to accomplish these?
Sure! Let's go step-by-step to solve these problems.
1) To determine the empirical formula of a compound, you need to find the simplest whole number ratio of the elements present in the compound.
Start by assuming you have 100g of the compound. This will make it easier to work with percentages.
In your compound, you have:
- Barium: 58.84g (58.84% of 100g)
- Sulfur: 13.74g (13.74% of 100g)
- Oxygen: 27.43g (27.43% of 100g)
Next, convert the mass of each element to moles. To do this, divide the mass of each element by its atomic weight. The atomic weights are:
- Barium (Ba): 137.33 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol
Now, calculate the number of moles:
- Moles of barium (Ba): 58.84g / 137.33 g/mol = 0.428 mol
- Moles of sulfur (S): 13.74g / 32.07 g/mol = 0.428 mol
- Moles of oxygen (O): 27.43g / 16.00 g/mol = 1.716 mol
The next step is to find the simplest whole number ratio of the elements. Divide each of the moles by the smallest value (0.428 mol in this case):
- Ba: 0.428 mol / 0.428 mol = 1
- S: 0.428 mol / 0.428 mol = 1
- O: 1.716 mol / 0.428 mol = 4
Therefore, the empirical formula of the compound is BaSO4.
2) Similar to the previous problem, start by assuming you have 100g of the compound. This will make it easier to work with percentages.
In your compound, you have:
- Aluminum: 1.271g (1.271% of 100g)
- Chlorine: 6.280g (6.280% of 100g)
Next, convert the mass of each element to moles. To do this, divide the mass of each element by its atomic weight:
- Aluminum (Al): 26.98 g/mol
- Chlorine (Cl): 35.45 g/mol
Calculating the number of moles:
- Moles of Aluminum (Al): 1.271g / 26.98 g/mol = 0.0471 mol
- Moles of Chlorine (Cl): 6.280g / 35.45 g/mol = 0.177 mol
Next, divide each of the moles by the smallest value (0.0471 mol in this case):
- Al: 0.0471 mol / 0.0471 mol = 1
- Cl: 0.177 mol / 0.0471 mol = 3.76
Rounding this to the nearest whole number, the empirical formula of aluminum chloride is AlCl4.
Sure! Let's break down the steps for solving these problems:
1) Determining the empirical formula of a compound:
Step 1: Convert the given percentages by mass into grams.
In this case, you are given percentages for barium (58.84%), sulfur (13.74%), and oxygen (27.43%). Assume you have 100 grams of the compound, so you can directly convert these percentages to grams.
Grams of barium (Ba) = 58.84 g
Grams of sulfur (S) = 13.74 g
Grams of oxygen (O) = 27.43 g
Step 2: Convert the grams of each element into moles.
To convert grams to moles, you need to divide the mass of each element by its molar mass. The molar mass of an element can be found on the periodic table.
Molar mass of barium (Ba) = 137.33 g/mol
Molar mass of sulfur (S) = 32.06 g/mol
Molar mass of oxygen (O) = 16.00 g/mol
Moles of barium (Ba) = 58.84 g / 137.33 g/mol
Moles of sulfur (S) = 13.74 g / 32.06 g/mol
Moles of oxygen (O) = 27.43 g / 16.00 g/mol
Step 3: Divide the number of moles of each element by the smallest number of moles obtained.
Dividing by the smallest number of moles will give you the simplest whole-number ratio between the different elements.
Divide the moles of each element by the smallest value:
Moles of barium (Ba) / smallest value
Moles of sulfur (S) / smallest value
Moles of oxygen (O) / smallest value
Step 4: Round off the obtained ratios (if necessary) to get the whole-number subscripts.
The ratios obtained from step 3 might not be whole numbers, so you need to round them off to the nearest whole number.
These whole numbers represent the subscript values, and they give you the empirical formula of the compound.
2) Calculating the empirical formula of a compound from experimental data:
Step 1: Convert the mass of each element into moles.
In this case, you are given the mass of aluminum (Al) metal (1.271 g) and the mass of aluminum chloride (AlCl3) produced (6.280 g).
Molar mass of aluminum (Al) = 26.98 g/mol
Moles of aluminum (Al) = 1.271 g / 26.98 g/mol
Moles of aluminum chloride (AlCl3) = 6.280 g / (mass of AlCl3)
Step 2: Determine the ratio between moles of aluminum and moles of chlorine (Cl).
Since aluminum chloride has one aluminum atom and three chlorine atoms, the mole ratio between aluminum and chlorine is 1:3.
Step 3: Write the empirical formula using the mole ratio.
The empirical formula of aluminum chloride is AlCl3. The subscript 3 indicates that there are three chlorine atoms for each aluminum atom.
I hope this helps you understand how to solve these types of problems!
Take a 100 g sample. That gives you
58.84 g Ba
13.74 g S
27.43 g O
Convert to moles. moles = grams/molar mass.
Then you want to find the ratio of the moles to each other with the smallest number being 1.00 (round if necessary but don't round too much). The easy way to do that is to divide the smallest number by itself; then divide all of the other numbers by the same small number.
For #2.
............Al + Cl2 = AlxCly
..........1.271........6.28
So you know Cl2 must be 6.28-l.271 = 5009g.
Then just follow the steps in problem 1.
Post your work if you have trouble.