Five mineral samples of equal mass of calcite, CaCO3 (MM 100.085) , had a total mass of 12.4 ± 0.1 g. What is the average mass of calcium in each sample?
(Assume that the relative uncertainties in atomic mass are small compared the uncertainty of the total mass.)
(Input the answer in terms of absolute uncertainty.)
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To find the average mass of calcium in each sample, we need to divide the total mass of calcium by the number of samples. We know that the mass of each sample is 12.4 ± 0.1 g.
First, let's calculate the molar mass of calcium. The molecular formula of calcite is CaCO3, which indicates that there is 1 atom of calcium in each molecule of calcite.
The molar mass of calcium is 40.08 g/mol.
Next, let's calculate the total number of moles of calcium in the 5 samples. We know that the mass of each sample is 12.4 g ± 0.1 g, so the total mass of calcium in the 5 samples is 12.4 g ± 0.1 g.
To find the number of moles, we use the formula:
Number of moles = Mass / Molar mass
Number of moles of calcium = (12.4 g ± 0.1 g) / (40.08 g/mol)
Using the uncertainty propagation formula for division:
Relative uncertainty of number of moles = (Relative uncertainty of mass) + (Relative uncertainty of molar mass)
Relative uncertainty of number of moles = (0.1 g / 12.4 g) + (0 g / 40.08 g/mol)
Relative uncertainty of number of moles = 0.0081 + 0
Relative uncertainty of number of moles = 0.0081
Therefore, the number of moles of calcium in the 5 samples is (12.4 ± 0.1) g / (40.08 ± 0) g/mol = 0.3084 ± 0.0081 mol.
To find the average mass of calcium in each sample, we divide the total mass of calcium by the number of samples:
Average mass of calcium = (12.4 g ± 0.1 g) / 5
Using the uncertainty propagation formula for division:
Absolute uncertainty of average mass of calcium = (Relative uncertainty of number of moles) * (Average mass of calcium)
Absolute uncertainty of average mass of calcium = 0.0081 * (12.4 g / 5)
Absolute uncertainty of average mass of calcium = 0.02004 g
Therefore, the average mass of calcium in each sample is 2.48 ± 0.02 g.
To find the average mass of calcium in each sample, we need to know the mass of calcium in one sample and then divide it by the number of samples.
First, let's calculate the molar mass of CaCO3. CaCO3 consists of one atom of calcium (Ca), one atom of carbon (C), and three atoms of oxygen (O). The atomic mass of calcium (Ca) is 40.08 g/mol, the atomic mass of carbon (C) is 12.01 g/mol, and the atomic mass of oxygen (O) is 16.00 g/mol.
The molar mass of CaCO3 can be calculated as follows:
Molar mass of CaCO3 = (1 * Ca) + (1 * C) + (3 * O)
= (1 * 40.08) + (1 * 12.01) + (3 * 16.00)
= 40.08 + 12.01 + 48.00
= 100.09 g/mol (rounding to five decimal places)
Now, to find the mass of calcium in one sample, we divide the molar mass of CaCO3 by the total number of atoms of calcium in one molecule of CaCO3. Since there is only one atom of calcium in one molecule of CaCO3, the mass of calcium in one sample would be equal to the molar mass of CaCO3.
Mass of calcium in one sample = 100.09 g/mol
Next, we need to find the average mass of calcium in each sample. This can be calculated by dividing the total mass of the samples by the number of samples.
Average mass of calcium in each sample = (Total mass of samples) / (Number of samples)
Given that the total mass of the samples is 12.4 ± 0.1 g, we can substitute this value into the formula.
Average mass of calcium in each sample = (12.4 ± 0.1 g) / (5)
Now, to calculate the absolute uncertainty, we can divide the absolute uncertainty of the total mass by the number of samples.
Absolute uncertainty of the average mass of calcium in each sample = (Absolute uncertainty of total mass) / (Number of samples)
= 0.1 g / 5
= 0.02 g
Therefore, the average mass of calcium in each sample is 2.48 ± 0.02 g (where the absolute uncertainty is ± 0.02 g).