A compound weighing 42 grams was found to contain 12 grams of magnesium,6 grams of carbon the rest oxgen.Determine the empirical formular.i got MgCO3.am i right or wrong
Mg = 24 g/mol
C = 12 g/mol
O = 16 g/mol
12/24 = 1/2 mol Mg
6/12 = 1/2 mol C
42 - 18 = 24 g O
24 g * 1 mol / 16 g = 3/2 mol O
yes, 1 1 3
MgCO3
thanku so much
Well, congrats for taking a stab at it! But I must say, you might need a little bit of a boost. So, let's work through it together, shall we?
First, let's calculate the amount of oxygen in the compound. Take the total compound weight of 42 grams and subtract the weights of magnesium (12 grams) and carbon (6 grams): 42 grams - 12 grams - 6 grams = 24 grams.
Now, we need to determine the ratio of elements in the compound. Divide the weight of each element by its atomic weight:
Magnesium: 12 grams ÷ 24.31 g/mol = 0.493
Carbon: 6 grams ÷ 12.01 g/mol = 0.499
Oxygen: 24 grams ÷ 16 g/mol = 1.5
Next, we need to get the simplest whole-number ratio of these values. Let's scale them up by multiplying them by 2:
Magnesium: 0.493 × 2 = 0.986
Carbon: 0.499 × 2 = 0.998
Oxygen: 1.5 × 2 = 3
Now, let's see if we can simplify these further:
Magnesium: 0.986 ÷ 0.986 ≈ 1
Carbon: 0.998 ÷ 0.986 ≈ 1
Oxygen: 3 ÷ 0.986 ≈ 3.04
Since we're looking for the simplest whole number ratio, let's approximate oxygen as 3. Therefore, the empirical formula of the compound is MgCO3. Well done! You got it right!
To determine the empirical formula of a compound, we need to find the simplest whole number ratio of elements present in the compound.
In this given compound, we have 12 grams of magnesium, 6 grams of carbon, and the remaining mass is oxygen.
To find the empirical formula, we should first convert each element's mass into moles by using their molar masses. The molar masses are:
- Magnesium (Mg): 24.31 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol
Converting the masses to moles:
Moles of Mg = 12 g / 24.31 g/mol = 0.49 mol
Moles of C = 6 g / 12.01 g/mol = 0.499 mol
Moles of O = (42 g - 12 g - 6 g) / 16.00 g/mol = 1.875 mol
Now, we need to divide each of these mole values by the smallest value to find the simplest whole number ratio.
Dividing by 0.49 (the smallest number):
Mg: 0.49 mol / 0.49 = 1
C: 0.499 mol / 0.49 ≈ 1.02
O: 1.875 mol / 0.49 ≈ 3.83
Since we are aiming for whole numbers, we need to round these values to the nearest whole number:
Mg: 1
C: 1
O: 4
Therefore, the empirical formula based on the given composition is MgCO₄, not MgCO₃.