When 0.500 g of copper(II) sulfate pentahydrate is heated at a temperature of 300°C for twenty minutes it loses some water of hydration. If the resulting sample weights 0.427 g then what is the formula of the resulting compound?
What equation(s) can I use to solve this?
How much water was in CuSO4.5H2O? That is
0.500g x (5mol H2O/1 mol CuSO4.5H2O) = xx g
0.500 g - 0.427 = yy g = mass H2O lost.
How much water still remains in the CuSO4.xH2O? That is xx g - yy g = zz g.
How much of that 0.427 g is CuSO4 with no water. 0.427 - water remaining so
0.427 - zz = w.
So you now have w g CuSO4 and zz g water remaining. Determine the ratio of each and the empirical formula from that. I get CuSO4.3H2O but check my thinking. Check my work.
Ah, the wonderful world of chemistry! Let's dive right into it, shall we?
To solve this problem, we can start by finding the mass of water lost during heating. We know that the initial sample weighs 0.500 g and the resulting sample weighs 0.427 g. So, the mass of water lost is 0.500 g - 0.427 g = 0.073 g.
Now, we need to determine the number of moles of water lost. To do this, we divide the mass of water lost by the molar mass of water. The molar mass of water is approximately 18 g/mol. So, the number of moles of water lost is 0.073 g / 18 g/mol = 0.004 moles.
To find the formula of the resulting compound, we can compare the number of moles of copper(II) sulfate pentahydrate before and after heating. Copper(II) sulfate pentahydrate has the chemical formula CuSO4·5H2O. From the formula, we see that there is one mole of copper(II) sulfate for every five moles of water.
Since the number of moles of water lost is 0.004, we divide it by 5 to find the number of moles of copper(II) sulfate. 0.004 moles / 5 = 0.0008 moles.
Now, we can write the formula of the resulting compound using the number of moles of copper(II) sulfate. Since the molar ratio is 1:1 between copper(II) sulfate and the resulting compound, the formula is simply CuSO4.
So, the formula of the resulting compound is CuSO4. And that, my friend, is how we clown scientists roll!
2008?
To solve this problem, you can use the equation for calculating the percentage of water of hydration in a compound:
% water of hydration = (mass loss / initial mass) * 100
You can also use the equation for calculating the number of moles:
moles = mass / molar mass
By using these equations, we can determine the formula of the resulting compound.
To solve this problem, we can use the concept of the law of conservation of mass. According to this law, the mass of the reactants in a chemical reaction should be equal to the mass of the products.
In this case, we start with copper(II) sulfate pentahydrate (CuSO4·5H2O) and heat it, causing it to lose water of hydration. The resulting compound will have a different formula, which we need to determine.
To begin, let's calculate the mass of water lost during heating. We start with 0.500 g of copper(II) sulfate pentahydrate and after heating, the remaining compound weighs 0.427 g. Therefore, the mass lost is:
Mass lost = Initial mass - Final mass
Mass lost = 0.500 g - 0.427 g
Mass lost = 0.073 g
Since water has a molar mass of approximately 18 g/mol, we can convert the mass lost to moles:
Moles of water lost = Mass lost / Molar mass of water
Moles of water lost = 0.073 g / 18 g/mol
Moles of water lost ≈ 0.004 moles
Copper(II) sulfate pentahydrate (CuSO4·5H2O) consists of copper(II) ions (Cu2+), sulfate ions (SO4^2-), and five water molecules (H2O). When it loses water of hydration, the remaining compound will have a different formula. Let's assume it has 'x' H2O molecules remaining.
Now, we can set up an equation using the principle of moles being conserved:
Number of moles of Cu + Number of moles of S + Number of moles of O + Number of moles of H2O lost = Number of moles of Cu + Number of moles of S + Number of moles of O + Number of moles of H2O remaining
We know that in the original copper(II) sulfate pentahydrate compound, there are five moles of water for every one mole of the compound.
Therefore, the number of moles of water remaining (x) is given by:
Number of moles of water remaining = 5 - (number of moles of water lost)
Number of moles of water remaining = 5 - 0.004
Number of moles of water remaining ≈ 4.996
Since there are four water molecules remaining, we can conclude that the formula of the resulting compound is CuSO4·4H2O.