A solution of a metal carbonate M2CO3, was prepared by dissolving 7.46 g of the anhydrous solid in water to give 1000cm^3 of solution. 25.0cm^3 of this solution reacted with 27.0cm^3 of 0.100 mol/dm^3 hydrochloric acid. Calculate the relative formula mass of M2CO3 and hence the relative atomic mass of the metal M
mass M2CO3 titrated = 7.46 x (25/1000) = approx 0.19 but you need a more accurate number for this and the all of the calculations that follow.
M2CO3 + 2HCl ==> CO2 + H2O + 2MCl
mols HCl = M x L = approx 0.0027
mols M2CO3 = 1/2*0.0027 = 0.00135
mols = grams/molar mass or
molar mass = g/mol = approx 0.19/0.00135 = about 140.
2M = 140-C-3*O = 140-12-48 = about 80 for 2M or approx 40 for M. My guess is K at 39.1
You had 7.46 grams to begin. You placed that in 1000 cc and pulled out 25 cc of that. So the amount in the 25 cc is all you titrated (you didn't titrate the entire 7.46g sample) so how much sample was in that 25 cc. That's
7.46 x 25/1000 =
I didn't understand the mass M2CO3 titrated = 7.46 x (25/1000) step. please, could you explain
Well, it seems like M2CO3 is being a bit shy and hiding behind all those numbers, doesn't it? Let's coax it out by doing some calculations!
First, let's calculate the number of moles of hydrochloric acid used:
25.0 cm^3 * 0.100 mol/dm^3 = 2.50 mmol (millimoles)
Since hydrochloric acid reacts in a 1:1 ratio with M2CO3, this tells us that 2.50 mmol of M2CO3 also reacted.
Now, let's calculate the number of moles of M2CO3 in the 1000 cm^3 solution:
7.46 g * (1 mol / molar mass of M2CO3) = 2.50 mmol
Since 2.50 mmol of M2CO3 reacted and that's equal to the amount of M2CO3 in the solution, we can conclude that the molar mass of M2CO3 is 7.46 g/mol.
But wait, we're not done yet! The relative formula mass of M2CO3 is double its molar mass since there are two "M" atoms in the formula. So, the relative formula mass of M2CO3 is 2 * 7.46 g/mol = 14.92 g/mol.
Now, since the relative formula mass of M2CO3 is equal to the relative atomic mass of M, we can conclude that the relative atomic mass of the metal M is 14.92 g/mol.
So, there you have it! The relative formula mass of M2CO3 is 14.92 g/mol, and the relative atomic mass of the metal M is also 14.92 g/mol. Clown approved! 🤡
To find the relative formula mass (RFM) of M2CO3 and the relative atomic mass of the metal M, we need to use stoichiometry and molar ratios based on the balanced chemical equation for the reaction between the metal carbonate and hydrochloric acid.
First, let's find the number of moles of hydrochloric acid used in the reaction:
Number of moles of HCl = concentration (mol/dm^3) × volume (dm^3)
Number of moles of HCl = 0.100 mol/dm^3 × 0.0270 dm^3
Number of moles of HCl = 0.0027 moles
Now, we need to find the number of moles of M2CO3 that reacted. Since the balanced chemical equation for the reaction is not provided, we assume a 1:1 molar ratio between the acid and the metal carbonate. Therefore, 0.0027 moles of HCl reacted with the same number of moles of M2CO3.
The concentration of M2CO3 in the original solution is given by:
Concentration of M2CO3 = mass (g) / volume (dm^3)
Concentration of M2CO3 = 7.46 g / 1.000 dm^3
Concentration of M2CO3 = 7.46 g/dm^3
Now, to find the number of moles of M2CO3 in the original solution, we use the molar mass of M2CO3:
Number of moles of M2CO3 = mass (g) / molar mass (g/mol)
Number of moles of M2CO3 = 7.46 g / (molar mass of M2CO3 g/mol)
Since we don't know the molar mass of M2CO3, we can't calculate it directly. However, we can find the relative formula mass (RFM) of M2CO3 by using the fact that 7.46 g of M2CO3 were dissolved in 1000 cm^3 (or 1.000 dm^3) of solution:
Concentration of M2CO3 (mol/dm^3) = moles of M2CO3 / volume (dm^3)
7.46 g/dm^3 = (7.46 g / RFM of M2CO3 g/mol) / 1.000 dm^3
Simplifying the equation above, we find:
RFM of M2CO3 = 7.46 g / 7.46 g/dm^3
Now, substituting the previously calculated number of moles of M2CO3 into the equation:
RFM of M2CO3 = 7.46 g / (concentration of M2CO3 in mol/dm^3)
Finally, once we determine the RFM of M2CO3, we can find the relative atomic mass of the metal M by dividing the RFM by 2, since there are two metal atoms in each molecule of M2CO3.
2HCl + M2CO3 >>> 2MCl + CO2 + H2O
So you reacted (25/1000)*7.46 grams of the carbonate, with .027*.1 moles of acid.
from the balanced equation, then the moles of M2CO3 is
1/2 * moles acid=1/2*.027*.1
figure that out.
Now you know (or can figure) the molmass of the carbonate
moles carbonate=above , then
moles carbonate=massused/molmass
or molmass=massused/moles carbonat
= 7.46g*(25/1000)/moles carbonate which you figured above.
Now haveing the mole mass of M2CO3, you can subtract the formula mass of CO3, and you are left with 2xAtomic mass M.
from that, determine the atomic mass of M.