Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# A student is given a sample of a green nickel sulfate hydrate. She weighs the sample in a dry covered crucible and obtains a mass of 22.326 g for the crucible, cover, and sample. earlier she had found that the crucible and cover weighed 21.244 g?

The mass of the hydrate was 1.082 g.

FULL QUESTION

A student is given a sample of a green nickel sulfate hydrate. She weighs the sample in a dry covered crucible and obtains a mass of 22.326 g for the crucible, cover, and sample. earlier she had found that the crucible and cover weighed 21.244 g.

She then heats the crucible to drive off the water of hydration, keeping the crucible at red heat for about 10 minutes with the cover slightly ajar. She then lets the crucible cool, and finds it has a lower mass; the crucible, cover and contents then weigh 21.841 g.

In the process the sample was converted to yellow anhydrous NiSO_4.

• What was the mass of the hydrate sample?
• What is the mass of the anhydrous NiSO_4?
• How much water was driven off?
• What is the percentage of water in the hydrate?
• How many grams of water would there be in 100.0 g hydrate? How many moles?

So, you know that the mass of the crucible, cover, and sample is 22.326 g. You can write

m_"crucible" + m_"cover" + m_"hydrate" = "22.326 g"

Moreover, you know that the mass of the crucible and cover is 21.244 g. This means that you have

overbrace(m_"crucible" + m_"cover")^(color(blue)("=21.244 g")) + m_"hydrate" = "22.326 g"

The initial mass of the sample was

m_"hydrate" = 22.326 - 21.244 = color(green)("1.082 g")

After you drive off the water of hydration, you find that the mass of the crucible, cover and anhydrous nickel sulfate is now 21.841 g.

Once again, you can write

overbrace(m_"crucible" + m_"cover")^(color(blue)("=21.244 g")) + m_"anhydrous" = "21.841 g"

This means that the mass of the anhydrous sample is

m_"anhydrous" = 21.841 - 21.244 = color(green)("0.597 g")

The difference between the initial weight of the sample and the anhydrous form is the water that you drove off.

m_"water" = m_"hydrate" - m_"anhydrous"

m_"water" = 1.082 - 0.597 = color(green)("0.485 g")

To get the percentage of water in the hydrate, you simply divide the mass of water by the mass of the hydrate and multiply this by 100.

"%water" = (0.485cancel("g"))/(1.082cancel("g")) * 100 = color(green)("44.82%")

This means that a 100-g sample of the hydrate would contain 44.82 g of water. To get the number of moles, use water's molar mass

44.82cancel("g") * "1 mole water"/(18.02cancel("g")) = color(green)("2.487 moles water")