What is the empirical formula for a compound of sodium and oxygen that contains 59% by mass?
I am guessing it would be NaO because the molar mass of Na is 22.99 g/mol
and 22.99 g/mol Na / 22.99 + 16 is approx 59%. Just wanted to make sure that is correct.
You're right
Well, sodium-oxygen compounds can be just full of "Naughty Oxides"! But in this case, you're absolutely right. The empirical formula for the compound containing 59% by mass of sodium and oxygen would be NaO. You've nailed it! Keep up the great work, chemistry whiz!
To determine the empirical formula of a compound, you need the ratio of the elements present in the compound. In this case, you are given that the compound contains 59% by mass of sodium (Na).
To calculate the empirical formula, you can assume that you have 100 grams of the compound. This means that 59 grams of the compound is sodium. The remaining mass (100 g - 59 g = 41 g) is oxygen (O).
Next, convert the masses of sodium and oxygen to moles using their respective molar masses. The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.
Moles of sodium (Na) = mass of sodium / molar mass of sodium = 59 g / 22.99 g/mol ≈ 2.57 mol
Moles of oxygen (O) = mass of oxygen / molar mass of oxygen = 41 g / 16.00 g/mol ≈ 2.56 mol
Now, divide the moles of each element by the smallest number of moles to get the simplest whole-number ratio of the elements.
Ratio of moles of sodium (Na) to moles of oxygen (O): 2.57 mol / 2.56 mol ≈ 1.00 : 1.00
So, the empirical formula of the compound is NaO, as you correctly guessed.
To determine the empirical formula for a compound, you need to find the smallest whole-number ratio of atoms present in the compound. In this case, we are given that the compound contains sodium (Na) and oxygen (O) and that it is 59% by mass.
To find the empirical formula, follow these steps:
1. Convert the given percentages to grams. Since the compound is 59% by mass, we can assume we have 100 grams of the compound. Therefore, we have 59 grams of sodium (Na) and 41 grams of oxygen (O).
2. Calculate the moles of each element. To do this, divide the number of grams of each element by their respective molar masses. The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.
- Moles of sodium (Na) = 59 g Na / 22.99 g/mol Na = 2.565 moles Na
- Moles of oxygen (O) = 41 g O / 16.00 g/mol O = 2.5625 moles O
3. Divide the moles of each element by the smallest number of moles calculated. In this case, the smallest number of moles is 2.5625 moles O.
- Sodium (Na): 2.565 moles Na / 2.5625 moles O = 1 mole Na (approximately)
- Oxygen (O): 2.5625 moles O / 2.5625 moles O = 1 mole O
The resulting ratio is approximately Na1O1, which simplifies to NaO. Therefore, your original guess was correct. The empirical formula for the compound containing sodium and oxygen is NaO.