Fasttasks 2 46 – The Troubleshooting Approximate

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1. Fasttasks 2 46 – The Troubleshooting Approximate Error
2. Fasttasks 2 46 – The Troubleshooting Approximate Square
3. Fasttasks 2 46 – The Troubleshooting Approximate Linear
4. Fasttasks 2 46 – The Troubleshooting Approximate The Number

Determine the formula of a hydrate
Fifteen Examples

TROUBLESHOOTING GUIDE The troubleshooting charts on the following pages have been developed to provide a systematic approach to handling problems. The charts provide troubleshooting information for each phase of tool operation, and list the suggested cor-rective action for each symptom beginning with the simplest of most likely solution first.

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
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)?

Photomatix pro 6 21. Solution:

1) Some preliminary comments: https://vescogalub1984.mystrikingly.com/blog/records-1-5-8-innovative-personal-database-training.

Ignore the water of hydration for a moment.

Gwen stefani the sweet escape itunes torrent. 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 Deliveries: a package tracker 1 1 2. 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:

Fasttasks 2 46 – The Troubleshooting Approximate Error

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:

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:

1.8707 mol / 0.11666 mol = 16.035

3) The formula is as follows:

Ga₂(SeO₄)₃⋅ 16H₂O

Example #11: http://cqhtydg.xtgem.com/Blog/__xtblog_entry/19081508-vehicle-simulator-crack-free-download#xt_blog. 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: Xplan 3 5 8 – track your projects.

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.

Fasttasks 2 46 – The Troubleshooting Approximate Square

(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.

Vip deluxe slots download. 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:

Fasttasks 2 46 – The Troubleshooting Approximate Linear

0.0224 mole / 2 = 0.0112 mol of carbonate

4) Determine the mass of 0.0112 mol of Na₂CO₃.

(0.0112 mol Na₂

Fasttasks 2 46 – The Troubleshooting Approximate The Number

CO₃) (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