Chromium (II) nitrate forms a hydrate that is 40.50% water by mass. What is its chemical formula?
I just need to figure out how to do it!
Cr(NO3)2.xH2O. You want to determine the value of x.
Take a 100 g sample which will give you
40.5 g H2O and
59.5 g Cr(NO3)2
Convert to moles.
40.5/molar mass H2O = ??
59.5/molar mass Cr(NO3)2
Now find the ratio of H2O molecules to 1 molecule Cr(NO3)2.
Check your numbers; this doesn't come out as well as I think it should. Check my work; perhaps I made an error.
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To determine the chemical formula of Chromium (II) nitrate hydrate, we need to break down the information provided step-by-step and use some chemical knowledge.
1. Let's start by writing the chemical formula for Chromium (II) nitrate without the water of hydration. The roman numeral (II) indicates that Chromium has a +2 charge. Nitrate is a polyatomic ion with a -1 charge. Since Chromium has a +2 charge and Nitrate has a -1 charge, we need two Nitrate ions to balance the charge of one Chromium ion. Therefore, the formula for Chromium (II) nitrate is Cr(NO3)2.
2. Now, we need to take into account that the hydrate contains 40.50% water by mass. This means that 40.50 g of the hydrate is water, and the remaining mass is the anhydrous Chromium (II) nitrate.
3. We need to calculate the mass of the anhydrous Chromium (II) nitrate. Since the hydrate is 40.50% water, the remaining mass is 100% - 40.50% = 59.50% anhydrous Chromium (II) nitrate.
4. Let's assume we have 100 g of the hydrate. This means that 40.50 g is water and 59.50 g is anhydrous Chromium (II) nitrate. The molar mass of water (H2O) is 18.015 g/mol, and the molar mass of Chromium (II) nitrate (Cr(NO3)2) is 166.01 g/mol.
5. We can calculate the number of moles of water using its mass and molar mass. The moles of water = mass of water / molar mass of water = 40.50 g / 18.015 g/mol.
6. Similarly, we can calculate the number of moles of anhydrous Chromium (II) nitrate. The moles of anhydrous Chromium (II) nitrate = mass of anhydrous Chromium (II) nitrate / molar mass of Chromium (II) nitrate = 59.50 g / 166.01 g/mol.
7. Now, we can find the ratio of moles between water and anhydrous Chromium (II) nitrate. Divide the moles of water by the moles of anhydrous Chromium (II) nitrate and simplify if necessary.
8. The simplified ratio of moles will give us the ratio of subscripts in the chemical formula. For example, if the ratio is water:Chromium (II) nitrate = 2:3, the chemical formula would be Cr(NO3)2*2H2O, indicating that there are two water molecules associated with each formula unit of Chromium (II) nitrate.
By following these steps, you should be able to determine the chemical formula of the hydrate.
To determine the chemical formula of chromium (II) nitrate hydrate, we need to understand the composition of the compound and its hydrate.
First, let's focus on chromium (II) nitrate. The formula for chromium (II) can be determined by considering the charge on the chromium ion (Cr2+). Since nitrate (NO3-) has a charge of -1, one Cr2+ ion will need two NO3- ions to balance the charges. Thus, the formula for chromium (II) nitrate is Cr(NO3)2.
Now let's consider the hydrate. A hydrate is a compound that contains a certain amount of water molecules chemically attached to its structure. In this case, the hydrate is reported to be 40.50% water by mass.
To calculate the number of water molecules per formula unit of the hydrate, we need to convert the percentage mass of water to a molar ratio. This can be done by assuming a 100 gram sample, as percentages are based on 100g.
Given that the hydrate is 40.50% water, we can deduce that there are 40.50g of water in a 100g sample. To convert this to moles, we need to use the molar mass of water, which is 18.015 g/mol.
So, the number of moles of water in the 100g sample can be calculated using the following equation:
moles of water = mass of water (g) / molar mass of water (g/mol)
moles of water = 40.50 g / 18.015 g/mol = 2.247 mol
Now, to determine the water-to-solute ratio, we compare the number of moles of water to the moles of chromium (II) nitrate.
From the formula Cr(NO3)2, we can see that each formula unit has one chromium ion (Cr2+) and two nitrate ions (NO3-). Therefore, the ratio of moles of water to moles of chromium (II) nitrate is:
moles of water : moles of Cr(NO3)2 = 2.247 : 1
To simplify this ratio, we multiply both sides by the reciprocal of the number of moles of Cr(NO3)2:
1 / 1 = 2.247 / moles of Cr(NO3)2
moles of Cr(NO3)2 = 2.247
Since we cannot have fractional moles, we need to multiply the ratio by the smallest whole number that gives us whole numbers of moles. In this case, multiplying by 2 gives us whole numbers:
2 moles of water : 2 moles of Cr(NO3)2 = 4.494 : 2
moles of Cr(NO3)2 = 4.494
This means that for every 4.494 moles of water, there are 2 moles of chromium (II) nitrate. Note that since we are dealing with hydrates, it is common practice to simplify the ratio and express it as the lowest whole number ratio.
Therefore, the chemical formula for chromium (II) nitrate hydrate is Cr(NO3)2 · 4.5H2O (rounded to one decimal place).