Calculate the amount of Substance in each of the following
(a)120g of Trioxocarbonate(IV)ion CO3.
(b) 31.5g of trioxocarbonate(V) acid HNO3.
(C)123g of copper(II) trioxocarbonate(IV).
Relative atomic mass
C=12
O=16
N=17
Cu=63
Ronald, I don't get it.
First, the IUPAC names are carbonate ion for A, carbonic acid for B (which btw, is H2CO3 and not HNO3) and C is copper(II) carbonate.
If I read the problem right, there are 123 g CuCO3 in 123 g of C. There are 31.5 g H2CO3 in 31.5 g H2CO3 in B and there are 120 g carbonate ion in A.
Surely that isn't the question.
Second, oobleck showed you how to solve one of the three. The others are done the same way and I'm sure your calculator works the same way as his, assuming that he has interpreted the problem correctly.
I really need the answer urgently please
Thanks oobleck
I need the answer to B and C too
CO3: C = 12 O3 = 48
so C = 12/(12+48) * 120g = 24g
and so on
To calculate the amount of substance in each of the given compounds, we need to use the formula:
Amount of Substance (in moles) = Mass (in grams) / Molar mass (in grams per mole)
First, we need to find the molar mass of each compound by adding up the atomic masses of each element present in the compound.
(a) Trioxocarbonate(IV) ion, CO3:
Molar mass of C = 12 g/mol
Molar mass of O = 16 g/mol (x3 since there are three oxygen atoms in CO3)
Molar mass of CO3 = (12 g/mol) + (16 g/mol x 3) = 48 g/mol
Next, we can calculate the amount of substance:
Amount of Substance = 120 g / 48 g/mol = 2.5 mol
Therefore, there are 2.5 moles of Trioxocarbonate(IV) ion in 120 grams.
(b) Trioxocarbonate(V) acid, HNO3:
Molar mass of H = 1 g/mol
Molar mass of N = 14 g/mol
Molar mass of O = 16 g/mol (x3 since there are three oxygen atoms in NO3)
Molar mass of HNO3 = (1 g/mol) + (14 g/mol) + (16 g/mol x 3) = 63 g/mol
Amount of Substance = 31.5 g / 63 g/mol = 0.5 mol
Therefore, there are 0.5 moles of Trioxocarbonate(V) acid in 31.5 grams.
(c) Copper(II) trioxocarbonate(IV):
Molar mass of Cu = 63 g/mol
Molar mass of C = 12 g/mol
Molar mass of O = 16 g/mol (x3 since there are three oxygen atoms in CO3)
Molar mass of CuCO3 = (63 g/mol) + (12 g/mol) + (16 g/mol x 3) = 123 g/mol
Amount of Substance = 123 g / 123 g/mol = 1 mol
Therefore, there is 1 mole of Copper(II) trioxocarbonate(IV) in 123 grams.
By using the given relative atomic masses and the formula for calculating amount of substance, we can determine the moles of each compound based on their respective masses.