Determine the formula of a hydrate
Fifteen Examples

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Hydrate problems #1 - 10	Calculate empirical formula when given mass data
Hydrate problems #11 - 25	Calculate empirical formula when given percent composition data
A list of all the problems	Determine identity of an element from a binary formula and a percent composition
Mole Table of Contents	Determine identity of an element from a binary formula and mass data

Example #1: A 15.67 g sample of a hydrate of magnesium carbonate was heated, without decomposing the carbonate, to drive off the water. The mass was reduced to 7.58 g. What is the formula of the hydrate?

Solution:

1) Determine mass of water driven off:

15.67 − 7.58 = 8.09 g of water

2) Determine moles of MgCO₃ and water:

MgCO₃ ⇒ 7.58 g / 84.313 g/mol = 0.0899 mol
H₂O ⇒ 8.09 g / 18.015 g/mol = 0.449 mol

3) Find a whole number molar ratio:

MgCO₃ ⇒ 0.0899 mol / 0.0899 mol = 1

Fasttasks 2 46 – The Troubleshooting Approach Emphasizes

H₂O ⇒ 0.449 mol / 0.0899 mol = 5

MgCO₃· 5H₂O

Example #2: A hydrate of Na₂CO₃ has a mass of 4.31 g before heating. After heating, the mass of the anhydrous compound is found to be 3.22 g. Determine the formula of the hydrate and then write out the name of the hydrate.

Solution:

1) Determine mass of water driven off:

4.31 − 3.22 = 1.09 g of water

2) Determine moles of Na₂CO₃ and water:

Na₂CO₃ ⇒ 3.22 g / 105.988 g/mol = 0.0304 mol
H₂O ⇒ 1.09 g / 18.015 g/mol = 0.0605 mol

3) Find a whole number molar ratio:

Na₂CO₃ ⇒ 0.0304 mol / 0.0304 mol = 1
H₂O ⇒ 0.0605 mol / 0.0304 mol = 2

Na₂CO₃· 2H₂O

sodium carbonate dihydrate

Comment: sodium carbonate forms three hydrates and the above is not one of them. This is a problem probably crafted so that you cannot look up possible answers via the InterTubez®. Just sayin'.

Example #3: When you react 3.9267 grams of Na₂CO₃· nH₂O with excess HCl(aq), 0.6039 grams of a gas is given off. What is the number of water molecules bonded to Na₂CO₃ (value of n)?

Solution:

1) Some preliminary comments:

Ignore the water of hydration for a moment.

Na₂CO₃(s) + 2HCl(aq) ---> 2NaCl(aq) + CO₂(g) + H₂O(ℓ)

The key is that there is a 1:1 molar ratio between Na₂CO₃ and CO₂. (Also, note that we assume that the gas is pure CO₂ and that there is no water vapor whatsoever. All of the water stays as a liquid. We also assume that no CO₂ dissolves in the water.)

2) Determine moles of CO₂:

0.6039 g / 44.009 g/mol = 0.013722 mol of CO₂

3) Use the 1:1 molar ratio referenced above:

This means that the HCl reacted with 0.013722 mole of sodium carbonate.

4) How many grams of Na₂CO₃ is that?

0.013722 mol times 105.988 g/mol = 1.4544 g

5) Determine grams, then moles of water

3.9267 g − 1.4544 g = 2.4723 g of water

2.4723 g / 18.015 g/mol = 0.13724 mol of water

6) For every one Na₂CO₃, how many waters are there?

0.13724 mol / 0.013722 mol = 10

Na₂CO₃· 10H₂O

Comment: this is one of the three sodium carbonate hydrates that exist.

Example #4: If 1.951 g BaCl₂· nH₂O yields 1.864 g of anhydrous BaSO₄ after treatment with sulfuric acid, calculate n.

Solution:

1) Calculate mass of Ba in BaSO₄:

(1.864 g) (137.33 g/mol / 233.39 g/mol) = 1.0968 g

2) Calculate mass of anhydrous BaCl₂ that contains 1.0968 g of Ba:

1.0968 g is to 137.33 g/mol as x is to 208.236 g/mol

x = 1.663 g

3) Calculate mass of water in original sample:

1.951 g − 1.663 g = 0.288 g

4) Calculate moles of anhydrous BaCl₂ and water:

1.663 g / 208.236 g/mol = 0.0080 mol
0.288 g / 18.015 g/mol = 0.0160 mol

5) Express the above ratio in small whole numbers with BaCl₂ set to a value of one:

BaCl₂ ---> 0.0080 mol / 0.0080 mol = 1
H₂O ---> 0.0160 mol/ 0.0080 mol = 2

BaCl₂· 2H₂O

Example #5: Given that the molar mass of Na₂SO₄· nH₂O is 322.1 g/mol, calculate the value of n.

Solution:

1) The molar mass of anhydrous Na₂SO₄ is:

142.041 g/mol

2) The mass of water in one mole of the hydrate is:

322.1 g − 142.041 g = 180.059 g

3) Determine moles of water:

180.059 g / 18.0 g/mol = 10 mol

4) Write the formula:

Na₂SO₄· 10H₂O

Example #6: 4.92 g of hydrated magnesium sulphate crystals (MgSO₄⋅ nH₂O) gave 2.40 g of anhydrous magnesium sulfate on heating to a constant mass. Determine the value of n.

Solution:

1) Mass of water:

4.92 g − 2.40 g = 2.52 g

2) Moles of water:

2.52 g / 18.0 g/mol = 0.14 mol

3) Moles of anhydrous MgSO₄:

2.40 g / 120.4 g/mol = 0.020 mol

4) Determine smallest whole-number ratio:

MgSO₄ ---> 0.020 / 0.020 = 1
H₂O ---> 0.14 / 0.020 = 7

MgSO₄⋅ 7H₂O

Over the time I put the above examples (and the problems) together, I never addressed the following type of problem, one in which the full formula of the hydrate is known and you are asked how much anhydrate remains after driving off the water. When I realized the lack, I put three examples here (as opposed to their own separate file) and included another in the problems.

There are three different ways you can approach this type of problem. What follows is an example of each way.

Example #7: A 241.3 gram sample of CoCl₂⋅ 2H₂O is heated to dryness. Find the mass of anhydrous salt remaining.

Solution using percent water:

1) Determine the percentage of water in cobalt(II) chloride dihydrate:

CoCl₂⋅ 2H₂O ---> 165.8686 g (in one mole)
mass of two moles of water ---> 36.0296 g

decimal percent of water in the hydrate ---> 36.0296 g / 165.8686 g = 0.217218

2) Determine mass of water in 241.3 g of the hydrate:

(241.3 g) (0.217218) = 52.4147 g

3) Determine mass of anhydrous salt remaining after heating to dryness:

241.3 g − 52.4147 g = 188.9 g (to four sig figs)

Example #8: A 2.56 g sample of ZnSO₄⋅ 7H₂O is heated to dryness. Determine the anhydrous mass remaining after all the water has been driven off.

Solution using percent anhydrate:

1) Determine percentage of anhydrous zinc sulfate in zinc sulfate heptahydrate:

molar mass of ZnSO₄⋅ 7H₂O ---> 287.5446 g
molar mass of ZnSO₄ ---> 161.441 g

decimal percent of anhydrate in the hydrate ---> 161.441 / 287.5446 = 0.561447 (keep some extra digits)

2) Determine mass of anhydrate in 2.56 g of the hydrate:

(2.56 g) (0.561447) = 1.4373 g

To three sig figs, this is 1.44 g

Example #9: If 29.0 g of MgSO₄⋅ 7H₂O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

Solution using 1:1 molar ratio:

1) Convert the 29.0 g of magnesium sulfate heptahydrate to moles:

Fasttasks 2 46 – The Troubleshooting Approach Focuses

29.0 g / 246.4696 g/mol = 0.11766157 mol

2) Note that, after heating, you end up with anhydrous magnesium sulfate, MgSO₄ and you will have remaining the same number of moles of the anhydrous product as you had moles of the hydrated reactant. In other words, there is a 1:1 molar ratio between the hydrated compound used and the anhydrous compound produced.

0.11766157 mol of MgSO₄ is produced

3) Determine grams of anhydrate produced:

(0.11766157 mol) (120.366 g/mol) = 14.16245 g

To three sig figs, this is 14.2 g

Example #10: What is the formula of the hydrate formed when 66.3 g of Ga₂(SeO₄)₃ combines with 33.7 g of H₂O?

Solution:

1) Determine moles of each substance present:

Ga₂(SeO₄)₃ ---> 66.3 g / 568.314 g/mol = 0.11666 mol
H₂O ---> 33.7 g / 18.015 g/mol = 1.8707 mol

2) We want to know who many moles of water are present when one mole of Ga₂(SeO₄)₃ is present:

Fasttasks 2 46 – The Troubleshooting Approach Suggests

1.8707 mol / 0.11666 mol = 16.035

3) The formula is as follows:

Ga₂(SeO₄)₃⋅ 16H₂O

Example #11: A student determined that the percent of water in a hydrate was 25.3%. The formula of the anhydrous compound was determined to be CuSO₄. Calculate the formula of the hydrated compound.

Solution:

1) Assume 100. g of the hydrate is present. That means this:

CuSO₄ ---> 74.7 g
H₂O ---> 25.3 g

2) Change to moles:

CuSO₄ ---> 74.7 g / 159.607 g/mol = 0.468 mol
H₂O ---> 25.3 g / 18.0 g/mol = 1.406 mol

3) Divide by the smallest:

CuSO₄ ---> 0.468 mol / 0.468 mol = 1
H₂O ---> 1.406 mol / 0.468 mol = 3

4) Formula:

CuSO₄· 3H₂O

The most common CuSO₄ hydrate is the pentahydrate, but the trihydrate does exist.

Example #12: 0.572 grams of a hydrate is heated to dryness, ending with 0.498 grams of anhydrous compound. What is the percentage of water by mass in the hydrate?

Solution:

(0.572 g − 0.498 g) / 0.572 g = 0.129

0.129 * 100 = 12.9%

Example #13: 1.534 grams of BaCl₂· 2H₂O is heated to dryness. What will be the mass of BaCl₂(s) that remains?

Solution:

1) Determine moles of hydrate:

1.534 g / 244.2636 g /mol = 0.0062801 mol

2) 0.0062801 mol of the anhydrate remain:

(0.0062801 mol) (208.233 g/mol) = 1.308 g

Example #14: 31.0 g of MgSO₄⋅ 7H₂O is thoroughly heated. What mass of anhydrous magnesium sulfate will remain?

Solution:

1) Here's a dimensional analysis solution:

1 mol	1 mol	120.3676 g
31.0 g x	–––––––––	x	–––––	x	–––––––––	= 15.1 g (to three sig figs)
246.4746 g	1 mol	1 mol

2) Explanation of the steps:

(a) 31.0 g of MgSO₄⋅ 7H₂O is divided by the molar mass of MgSO₄⋅ 7H₂O to give moles of MgSO₄⋅ 7H₂O.

(b) The molar ratio of 1 to 1 is derived from this equation:

MgSO₄⋅ 7H₂O(s) ---> MgSO₄(s) + 7H₂O(ℓ)

For every one mole of MgSO₄⋅ 7H₂O heated, one mole of anhydrous MgSO₄ is left behind when all the water is driven off.

(c) The moles of MgSO₄ are multipled by the molar mass of MgSO₄ to give grams of MgSO₄ (the answer).

Example #15: When a hydrate of Na₂CO₃ is heated until all the water is removed, it loses 54.3 percent of its mass. Determine the formula of the hydrate.

Solution #1:

1) Let us assume one mole of the hydrated Na₂CO₃ is present. The molar mass of anhydrous Na₂CO₃ is 105.988 g/mol.

2) The hydrate sample lost 54.3% of its mass (all water) to arrive at 105.988 g. This means that the 105.988 g is 45.7% of the total mass.

3) We can now write a ratio and proportion:

105.988 g	x
–––––––	=	–––––––
45.7	100

x = 231.921 g (this is the molr mass of the hydrate since one mole of it was present at the start)

4) Determine mass, then moles of water:

231.921 − 105.988 = 125.933 g

125.933 g / 18.015 g = 6.99

5) The formula:

Na₂CO₃· 7H₂O

Solution #2:

1) Assume 100 g of Na₂CO₃· nH₂O is present. Therefore, of the 100 grams:

Na₂CO₃ = 45.7 g
H₂O = 54.3 g

2) Convert mass to moles:

Na₂CO₃ ---> 45.7 g / 105.988 g/mol = 0.43118 mol
H₂O ---> 54.3 g / 18.015 g/mol = 3.014155 mol

3) Set up a ratio and proportion:

0.43118 mol	1
–––––––––––	=	–––––––
3.014155 mol	n

n = 6.99

4) The formula:

Na₂CO₃· 7H₂O

Bonus Example: 3.20 g of hydrated sodium carbonate, Na₂CO₃⋅ nH₂O was dissolved in water and the resulting solution was titrated against 1.00 mol dm¯³ hydrochloric acid. 22.4 cm³ of the acid was required. What is the value of n?

Solution:

1) Sodium carbonate dissolves in water as follows:

Na₂CO₃⋅ nH₂O(s) ---> 2Na⁺(aq) + CO₃²¯(aq) + nH₂O(ℓ)

2) The addition of HCl will drive all of the CO₃²¯ ion to form CO₂ gas. One mole of carbonate ion will produce n moles of water.

CO₃²¯ + 2H⁺ ---> CO₂(g) + H₂O(ℓ)

3) Determine moles of HCl and from that moles of carbonate:

MV = moles

(1.00 mol/L) (0.0224 L) = 0.0224 mole of HCl

Two moles of HCl react for every one mole of carbonate. Therefore:

0.0224 mole / 2 = 0.0112 mol of carbonate

4) Determine the mass of 0.0112 mol of Na₂CO₃.

(0.0112 mol Na₂CO₃) (105.988 g Na / 1 mol) = 1.187 g Na₂CO₃

This is how many moles of anhydrous sodium carbonate dissolved.

5) Mass of hydrated salt − mass of anhydrous salt = mass of water

3.20 g − 1.187 g = 2.013 g H₂O

6) Convert to moles of water

2.013 g H₂O x (1 mol/18.015 g) = 0.11174 mol H₂O

7) Determine smallest whole-number ratio between sodium carbonate and water:

Na₂CO₃: 0.0112 mol / 0.0112 = 1
H₂O: 0.11174 mol / 0.0112 = 9.97 = 10

Na₂CO₃⋅ 10H₂O

Hydrate problems #1 - 10	Calculate empirical formula when given mass data
Hydrate problems #11 - 25	Calculate empirical formula when given percent composition data
A list of all the problems	Determine identity of an element from a binary formula and a percent composition
Mole Table of Contents	Determine identity of an element from a binary formula and mass data

Technical Questions

Do I need to wear leathers? One piece or two-piece? What does 2Fast require for gear at their schools & trackdays?

We require a one or two piece suit with protective armor only; while we highly recommend such suits be leather, we do not require that only leather suits be worn. Non-leather suits such as Joe Rocket or Aerostitch brand riding suits are acceptable. Additionally, 2Fast prefers that your 2-piece suit, no matter what it is made of, zips the pant and jacket together; however, we do not require that the pant and jacket zip together. Jeans or non-motorcycle wear (ie. motocross) are not allowed. To be very clear, your gear – whether leather or not – must have protective armor in the elbows, shoulders, and knees at a minimum.

We also require a DOT approved motorcycle helmet in good condition, leather boots that are over the ankle, and leather riding gloves.

Riders participate at their own risk, and the choice of what you choose to wear or not wear is your own. No matter what you choose to wear, and whether it zips or doesn’t, let it be known that 2Fast is a proponent of wearing the safest, most protective gear available. For this reason, we believe that every rider on the track and on public streets should be wearing full leather protective gear head to toe, and if the protective wear is a 2-piece suit, then it should zip together.

Motorcycle riding is a dangerous sport: dress accordingly. Don’t cut corners on gear.

What do I need to do to my bike in preparation for a 2Fast track day?

Make sure your bike runs well, is clean of debris, and is 100% free of oil leaks. Check your tires for tread wear and condition (75% tread life remaining or better). Tape over the glass/plastic on your lights and turn signals, as well as the glass on your mirrors. Either disable or tape over your rear brake light completely; we don’t want to be able to see it working on the track, as it can be distracting. Be sure your throttle returns to a closed position quickly (snaps back) when rolled open and then released. Any bikes with throttles that do not return quickly or are “sticky” will not be allowed on the track. Also check your tire pressures, your brakes, and be sure to have a full tank of gas in the bike along with a full gas can or two to use during the day. Check oil levels, brake levers and pads, throttle cables and operation, etc. No safety wiring or coolant change required!

How do I know if my tires will pass inspection and be suitable for the track?

First, do you have 75% or better tread left on both front and rear? This is the first test, and if you are in doubt, get new tires. The track will wear down tires many times quicker than a street ride will, so have tires with plenty of life in them. Second, what’s the age on the tires? If your bike hasn’t been ridden in a year, for instance, and the tires feel hard as a rock, get new tires. Bottom line is this: don’t go cheap on tires. They’re the difference between good traction and bad, which can be the difference between a fun day… or a day with a crash.

Will there be tires available for sale at the track, and/or a tire changing service?

Yes. Tires and tire changing service is available through our track day vendors. Barry Wressell of KFG Racing provides Dunlop tires and also provides suspension services. Tom Young of Competition Motorsports provides Pirelli tires and also has various supplies for track day riders. Both are in attendance at each of our events offering tires for sale – as well as many parts and accessories. If you’d like to make arrangements with either Barry or Tom, please contact either one using our trackside page.

I ride a sportbike and I have sportbike tires. What tire pressures do you recommend for the track?

For riders using a performance/sport/track day tire, we suggest around 31-33psi in the front, and roughly 30-32psi in the rear. We would suggest a pound or two more in the front tire. HOWEVER, each tire manufacturer has different pressure requirements and recommendations in order to make a particular tire perform at its best. Now add weather/temperature variances: hot days vs. cold. Thus, our best suggestion is to talk to one of our trackside tire vendors upon your arrival and get their thoughts on tire pressures for your tire brand and model, and the weather on that day.

Must I tape my lights and turn signals? Mirrors?

Not necessarily. We require that your front headlight, tail light, and mirrors be inoperable or non-functioning for our event. Whether you disable your head and tail light via pulling the plug, removing a fuse, or taping them over, its up to you. We don’t want the light being emitted from them and thats our reason. In days of old, when lights were made from glass that could cause damage to riders or tires if broken out on the track, taping them prevented the shards from finding the track. With everything being composite plastic now a days, this is not necessary. In regards to the mirrors, you can tape or remove them. We want to remove your ability to look behind you and essentially take them away from you, so thats why taping or removing them are both permitted with 2Fast.