To convert from percent to decimal, divide the percent by 100. Dividing by 100 will shift the decimal two places to the left. So 25.0% = 25.0/100 =.250 Basic Equation: A% = A/100 (where A is a valid number) Decimal to Percent. Theory predicts that 46.59 g of sodium sulfate product is possible if the reaction proceeds perfectly and to completion. But the question states that the actual yield is only 37.91 g of sodium sulfate. With these two pieces of information, you can calculate the percent yield using the percent-yield formula. Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. For this calculator, the order of the numbers does not matter as we are simply dividing the difference between two numbers by the average of the two numbers. SlidePad 1.0.1 review A sometimes-irritating solution in search of a problem By Barnaby Page 29 March 2009. SlidePad is a sadly limited application unless you take a lot of small notes.
| Founded | 2009; 11 years ago |
|---|---|
| Headquarters | Palo Alto, California |
| Brian Riley CEO, co-founder, Andrew Ouelett Co-founder, Al Nordin Executive Advisor | |
| Products | Bicycle brakes |
| Website | Slidepadbrake.com |
Slidepad is a bicycle brake technology company located in Palo Alto, California.[1] It was co-founded by Brian Riley and Andrew Ouellet in 2009.[2]
Slidepad was started by Riley and Ouellet while students at Cal Poly, San Luis Obispo.[2] Ouellet received inspiration for the product after he crashed his bicycle when applying too much pressure on his front brakes, resulting in front wheel lockup.[2] In 2009, Riley and Ouellet entered their initial design in Cal Poly’s Business Plan competition and won 1st place.[3] In 2013 Alan Nordin, former president of Fallbrook Technologies' bicycle division, joined the company as an executive advisor.[4]
In 2011, Slidepad Technologies formed an agreement with a Taiwanese manufacturer to build a Slidepad braking system for OEM distribution.[5] Jamis Bicycles was the first bike-manufacturer to specify the technology on their 2013 models.[6]Stanford University and Jamis Bicycles currently use Slidepad technology.[5][7]
In November 2012, the company took a 40-day, 11,000 mile, 'Save Your Teeth Tour' across 90 bike shops from Palo Alto, California to New Jersey.[6] https://downefil385.weebly.com/anymp4-dvd-creator-6-1-52-download-free.html.
Slidepad is an Intelligent Brake Distribution (IBD) technology, aimed at making braking easier for novice or casual cyclists, integrates into V-brake systems to provide single-lever braking.[4] It modulates the front brake force in real time, based on the road surface and rider weight position, and avoids front wheel lockup accidents when applying the front brake.[1][8] It was designed to prevent riders from flipping over their handlebars when applying the front brake.[1] Once the brake pads make contact with the rear wheel, the Slidepad slides forward, which pulls a cable that is connected to the front brakes.[1] Similar to the anti-lock brake system in cars, the mechanism prevents the front wheel from locking, no matter how hard the brake is pressed or how slippery the road conditions are.[6][9]
Notice below how I do the first problem with some attention to using proper atomic weights, as well as keeping close to the proper number of significant figures. Then, notice how I get away from that (as well as being real consistent with units) in the following problems.
Notice also how it really doesn't make much of a difference. The trick is to know when to do that and it comes only via experience. Generally speaking, in empirical formula problems, C = 12, H = 1, O = 16 and S = 32 are sufficient.
There are times when using 12.011 or 1.008 will be necessary. If you hit a problem that just doesn't seem to be working out, go back and re-calculate with more precise atomic weights. These problems, however, are fairly uncommon.
For what it is worth, one piece of advice on rounding: don't round off on the moles if you see something like 2.33 or 4.665. That first one can be rendered as two and one-third (or seven thirds) and the second one as four and two-thirds (or fourteen thirds). In a situation like that, you would multiply by three to reach the smallest whole-number ratio rather than dividing by the smallest.
I know it's easy to say, harder to demonstrate. Some of the problems below involve this thirds issue. Look for a problem involving citric acid. Just be aware that rounding off too early and/or too much is a common problem in this type of problem.
Example #1: A compound is found to contain 50.05 % sulfur and 49.95 % oxygen by weight. What is the empirical formula for this compound? The molecular weight for this compound is 64.07 g/mol. What is its molecular formula?
Solution:
1) Assume 100 g of the compound is present. This changes the percents to grams:
S ⇒ 50.05 g
O ⇒ 49.95 g
2) Convert the masses to moles:
S ⇒ 50.05 g / 32.066 g/mol = 1.5608 mol
O ⇒ 49.95 g / 16.00 g/mol = 3.1212 mol
3) Divide by the lowest, seeking the smallest whole-number ratio:
S ⇒ 1.5608 / 1.5608 = 1
O ⇒ 3.1212 / 1.5608 = 2
4) Write the empirical formula:
SO2
5) Compute the 'empirical formula weight:'
32 + 16 + 16 = 64
6) Divide the molecule weight by the 'EFW:'
64.07 / 64 = 1
7) Use the scaling factor computed just above to determine the molecular formula:
SO2 times 1 gives SO2 for the molecular formula
Example #2: A compound is found to contain 64.80 % carbon, 13.62 % hydrogen, and 21.58 % oxygen by weight. What is the empirical formula for this compound? The molecular weight for this compound is 74.14 g/mol. What is its molecular formula?
Solution:
1) Assume 100 g of the compound is present. This changes the percents to grams:
C ⇒ 64.80 g
H ⇒ 13.62 g
O ⇒ 21.58 g
2) Convert the masses to moles:
C ⇒ 64.80 g / 12 = 5.4
H ⇒ 13.62 g / 1 = 13.62
O ⇒ 21.58 g / 16 = 1.349
3) Divide by the lowest, seeking the smallest whole-number ratio:
C ⇒ 5.4 / 1.349 = 4
H ⇒ 13.62 / 1.349 = 10
O ⇒ 1.349 / 1.349 = 1
4) Write the empirical formula:
C4H10O
5) Determine the molecular formula:
'EFW' ⇒ 48 + 10 + 16 = 7474.14 / 74 = 1
molecular formula = C4H10O
Example #3: A compound is found to contain 31.42 % sulfur, 31.35 % oxygen, and 37.23 % fluorine by weight. What is the empirical formula for this compound? The molecular weight for this compound is 102.2 g/mol. What is its molecular formula?
Solution:
1) Percents to mass, based on assuming 100 g of compound present:
S ⇒ 31.42 g
O ⇒ 31.35 g
F ⇒ 37.23 g
2) Calculate moles of each:
S ⇒ 0.982 mol
O ⇒ 1.96 mol
F ⇒ 1.96 mol
3) Smallest whole-number ratio:
S ⇒ 1
O ⇒ 2
F ⇒ 2
4) Write the empirical and molecular formula formula:
SO2F2'EFW' ⇒ 32 + 32 + 38 = 102 g
the empirical formula is also the molecular formula
Example #4: Ammonia reacts with phosphoric acid to form a compound that contains 28.2% nitrogen, 20.8% phosphorous, 8.1% hydrogen and 42.9% oxygen. Calculate the empirical formula of this compound.
Solution:
1) Masses:
N ⇒ 28.2 g
P ⇒ 20.8 g
O ⇒ 42.9 g
H ⇒ 8.1 g
2) Moles:
N ⇒ 2
P ⇒ 0.67
O ⇒ 2.68
H ⇒ 8
3) Lowest whole-number ratio:
N = 2 / 0.67 = 3
P = 0.67 / 0.67 = 1
O = 2.68 / 0.67 = 4
H = 8 / 0.67 = 12
4) Empirical formula:
N3H12PO4or
(NH4)3PO4
Although not asked for, the name of this compound is ammonium phosphate.
5) I would like to discuss my piece of advice (about thirds) at the top of the file using the moles data from the above problem.
N ⇒ 2 = 6/3
P ⇒ 0.67 = 2/3
O ⇒ 2.68 = 8/3
H ⇒ 8 = 24/3
6) Then, I multiply:
N ⇒ 6/3 times 3 = 6
P ⇒ 2/3 times 3 = 2
O ⇒ 8/3 times 3 = 8
H ⇒ 24/3 times 3 = 24
7) Notice how doing it this way introduces an extra factor of 2. We remove the extra factor of two to arrive at this ratio:
N ⇒ 3
P ⇒ 1
O ⇒ 4
H ⇒ 12
8) And we continue on. I really don't want you to think that the introduction of the extra factor of two damages this technique. There are times when changing everything to third-type fractions will make things easier.
As in this problem.
Example #5: A compound contains 57.54% C, 3.45% H, and 39.01% F. What is its empirical formula?
Solution:
1) Get mass, then moles:
C ⇒ 57.54 / 12.011 = 4.791
H ⇒ 3.45 / 1.0008 = 3.423
F ⇒ 39.01 / 19.00 = 2.053
2) Seek lowest whole-number ratio:
C ⇒ 4.791 / 2.053 = 2.33
H ⇒ 3.423 / 2.053 = 1.67
F ⇒ 2.053 / 2.053 = 1
3) The key here is to see that 2.33 is 2 and one-third or 7/3 and that 1.67 is 5/3. Therefore:
C ⇒ (7/3) x 3 = 7Empirical formula is C7H5F3
Example #6: Halothane is an anesthetic that is 12.17% C, 0.51% H, 40.48% Br, 17.96% Cl and 28.87% F by mass. What is the compound's molar mass if each molecule contains exactly one hydrogen atom? (Note: try and do this without a calculator.)
Solution:
Guess the formula as C2HBrClF3How'd I do that?
Divide each percent by the atomic weight of the element and you get this:
C = 1
H = 0.5
Br = 0.5
Cl = 0.5
F = 1.5
Multiply through by 2.
I think the key #1 in this problem is to see that the 12.17% of carbon will go to 12.17 g and that 12.17 / 12.011 is essentially equal to 1. Key #2 is to see that hydrogen would be 0.51 g / 1.0 g/mol = 0.5 mole and that you would need to multiply it by 2 to get to one H atom. That means there will have to be two carbons.
The other elements are attacked in the same way.
Example #7: A compound was found to contain 24.74% (by mass) potassium, 34.76% manganese, and 40.50% oxygen. Determine the empirical formula.
Solution:
I like the titles of each step used by the person who wrote this answer on Yahoo Answers.
1) Collect atomic mass:
Potassium (K) has 39.1 a.m.u.
Mnaganese (Mn) has 54.9 a.m.u.
Oxygen (O) has 16.0 a.m.u.
2) Calculate stoichiometric ratio:
K ⇒ 24.74 / 39.1 = 0.63
Mn ⇒ 34.76 / 54.9 = 0.63
O ⇒ 40.50 / 16.0 = 2.53
3) Find integer numbers on the basis of ratios:
K : Mn : O = 0.63 : 0.63 : 2.53 = 1 : 1 : 4
4) Write empirical formula:
KMnO4
Example #8: A mass spectrometer analysis finds that a molecule has a composition of 48% Cd, 20.8% C, 2.62% H, 27.8% O. Determine the empirical formula.
Solution:
1) Let us assume 100 g of the compound is present. This means:
48 g Cd, 20.8 g C, 2.62 g H, 27.8 g O
2) Let us determine moles present:
Cd ⇒ 48 g / 112.4 g/mol = 0.427 mol
C ⇒ 20.8 g / 12.011 g/mol = 1.732 mol
H ⇒ 2.62 g / 1.008 g/mol = 2.5992 mol
O ⇒ 27.8 g / 16.00 g/mol = 1.7375 mol
3) Divide through by lowest value:
Cd ⇒ 0.427 mol / 0.427 mol = 1
C ⇒ 1.732 mol / 0.427 mol = 4.06
H ⇒ 2.5992 mol / 0.427 mol = 6.09
O ⇒ 1.7375 mol / 0.427 mol = 4.07
4) Ignore the Cd and see a 4 : 6 : 4 ratio for C : H : O. Reduce it to 2 : 3 : 2. Therefore:
C2H3O2C2H3O2¯ is the acetate ion
5) Cadmium is divalent, so we can see the empirical formula as:
Cd(C2H3O2)2Notice how the molar ratio in the full formula for cadium acetate is 1 : 4 : 6 : 4
Example #9: A bromoalkane contains 35% carbon and 6.57% hydrogen by mass. Calculate the empirical formula of this bromoalkane.
Solution:
1) Assume 100 g of the compound is available:
C ⇒ 35 g
H ⇒ 6.57 g
Br ⇒ 58.43 g (from 100 minus 41.57)
2) Determine moles:
C ⇒ 35 g / 12 gmol = 2.917
H ⇒ 6.57 g / 1 g/mol = 6.57
Br ⇒ 58.43 g / 80 g/mol = 0.730375
3) Divide by smallest to seek lowest whole-number ratio:
C4H9Br
Example #10: A compound containing sodium, chlorine, and oxygen is 25.42% sodium by mass. A 3.25 g sample gives 4.33 x 1022 atoms of oxygen. What is the empirical formula?
Solution:
1) Percent oxygen in the sample:
4.33 x 1022 atoms divided by 6.022 x 1023 atoms/mol = 0.071903 mol0.071903 mol times 16.00 g/mol = 1.15045 g
1.15045 g / 3.25 g = 0.3540 = 35.40%
2) Percent chlorine:
100 minus (25.42 + 35.40) = 39.18%
3) Assume 100 g of the compound is present. This converts percents to grams. Determine moles:
Na ⇒ 25.42 g / 23.0 g/mol = 1.105
Cl ⇒ 39.18 g / 35.453 g/mol = 1.105
O ⇒ 35.40 g / 16.00 g/mol = 2.2125
4) Finish with lowest whole-number ratio:
Divide by 1.105 to get lowest whole-number ratio of 1 : 1 : 2NaClO2
Although not asked for, this is the formula for sodium chlorite. Blocs 3 3 0 cr2.
Example #11: Analysis of a compound containing only C and Br revealed that it contains 33.33% C atoms by number and has a molar mass of 515.46 g/mol. What is the molecular formula of this compound?
Solution:
1) '. . . 33.33% C atoms by number . . .' Since mole is a measure of how many (one mole = 6.022 x 1023 chemical entities), we know this:
C ⇒ 0.3333 mol
Br ⇒ 0.6667 mol
2) Let us determine the smallest whole-number ratio:
C ⇒ 0.3333 / 0.3333 = 1
Br ⇒ 0.6667 / 0.3333 = 2
3) The empirical formula is CBr2. Determine the molecular formula:
515.46 / 171.819 = 3C3Br6
Example #12: Chemical analysis shows that citric acid contains 37.51% C, 4.20% H, and 58.29% O. What is the empirical formula?
Solution:
1) We start by assuming 100 g of the compound is present. This turns the above percents into masses.
2) Calculate moles:
C ⇒ 37.51 / 12.011 = 3.123
H ⇒ 4.20 / 1.008 = 4.167
O ⇒ 58.29 / 15.999 = 3.643
3) Look for lowest whole-number ratio:
C ⇒ 3.123 / 3.123 = 1See that 1.334. That's one and one-third or 4/3. I'm going to multiply all three values by 3:
C ⇒ 1 x 3 = 3
H ⇒ 1.334 x 3 = 4
O ⇒ 1.166 x 3 = 3.5
See that 3.5? Let's now multiply through by 2.
C = 6
H = 8
O = 7
4) The empirical formula:
C6H8O7
When I found this question on Yahoo Answers, there was a wrong answer given:
C ⇒ 37.51/12 = 3.1258
H ⇒ 4.2/1 = 4.20
O ⇒ 58.29/16 = 3.6431
Mole proportion = CHO = Empirical formula.
Too much rounding off. Be very careful on rounding off or a problem like this citric acid one will trip you up. Learn to recognize that something like 1.334 should be thought of as 4/3, leading to multiplying through by three. Do not round 1.334 off to 1 or round off something like 2.667 to three. And certainly, do not round off like the wrong-answer person did. No no no!
Example #13: A compound is 19.3% Na, 26.9% S, and 53.8% O. Its formula mass is 238 g/mol. What is the molecular formula?
Solution:
http://gyoouan.xtgem.com/Blog/__xtblog_entry/18986780-nice-clipboard-1-6-128#xt_blog. 1) We start by assuming 100 g of the compound is present. This turns the above percents into masses.
2) Calculate moles:
Na ⇒ 19.3 / 23.00 = 0.84
S ⇒ 26.9 / 32.1 = 0.84
O ⇒ 53.8 / 16.00 = 3.36
3) Look for lowest whole-number ratio:
3.36 / 0.84 = 4 (I only did the one for oxygen. You should be able to figure out the other two values!)
4) The empirical formula:
NaSO4
4) The molecular formula:
238 / 119 = 2Na2S2O8
Example #14: In which I present a problem and solution stripped down to their essentials. Hope you enjoy it! C = 48.38%, H = 8.12%, O = 53.5%
Solution:
4.0281.2
2.4
1
12
24
10
C6H12O5
Interesting how you have a multiply by 10, then a divide by 2. You might ask: why not just multiply by 5? Well, you could, if you saw it. If you didn't, moving the decimal point to get whole numbers, then seeing the common factor gets you to the same place in a bit more educational way.
That being said, if you saw that a multiply by five works, then treat yourself to some ice cream!
Example #15: Nitroglycerin has the following percentage composition:
carbon: 15.87%, hydrogen: 2.22%, nitrogen: 18.50%, oxygen: 63.41%
Determine its empirical formula.
Solution:
The assumption that 100 g of the compound is present turns the above percents into grams.
1) Calculate moles (I'll ignore units):
carbon ---> 15.87 / 12.01 = 1.321
hydrogen ---> 2.22/1.01 = 2.198
nitrogen ---> 18.50/14.01 = 1.320
oxygen ---> 63.41/16.0 = 3.963
2) Seek lowest whole-number ratio:
C ---> 1.321 / 1.32 = 1
H ---> 2.198 / 1.32 = 1.66
N ---> 1.32 / 1.32 = 1
O ---> 3.963 / 1.32 = 3
The key is the 1.66 which you do not round off to two. Think of it as 5/3.
3) Multiply everything by 3:
C ---> 1 x 3 = 3
H ---> 5/3 x 3 = 5
N ---> 1 x 3 = 3
O ---> 3 x 3 = 9
4) Empirical formula is:
C3H5N3O9
Example #16: Insulin contains 3.4% sulphur. Calculate minimum molecular mass of insulin.
Solution:
1) Assume 100 g of insulin is present.
3.4% of that 100 grams is sulfur. Therefore, 3.4 g of sulfur is present.
2) Determine how many moles of sulfur are are in 3.4 g of sulfur:
3.4 g / 32.065 g/mol = 0.106035 mol
3) Assume one mole of insulin contains one mole of sulfur:
| 0.106035 | 100 g | |
| ––––––– | = | ––––––– |
| 1 | x |
0.106035x = 100 g
x = 943 g
Example #17: Two metallic oxides contain 27.6% and 30% oxygen in them respectively. If the formula of the first oxide is M3O4, then, what will be the formula of the second?
Solution:
I will reproduce the answer given on Yahoo Answers:
Here you express everything on a per 100 g basis. The first oxide contains 27.6 g O or 27.6/16 = 1.725 moles of O and metal will be 100 - 27.6 = 72.4 g. Now the formula given is M3Do similar calculations for the second one, 30 g O = 30/16 = 1.875 moles reacting with 100-30 = 70 g metal. So the moles of metal will be 70/56 = 1.25 molesso the ratio of metal to oxygen is 1.25:1.875, divide by the smaller number which is 1.25, you get 1:1.5, you need to get to whole numbers, so it will be 2:3, therefore the formula will be M2O3
Example #18: What formula yields 36.8% nitrogen in a nitrogen oxide?
Solution:
1) Write this:
| 14N | |
| 0.368 = | ––––––––––– |
| 14N + 16O |
Where N = the number of nitrogen atoms and O = the number of oxygen atoms
2) Cross multiply:
5.152N + 5.888O = 14N
3) Collect like terms:
5.888O = 8.848N
4) Divide through by smallest:
O = 1.5NWhen N = 2, O = 3
5) The formula:N2O3
6) Another way to think of it:
1.5N = O3N = 2O
| 3N | 2O | |
| ––––––– | = | ––––––– |
| (2) (3) | (2) (3) |
| N | O | |
| ––––––– | = | ––––––– |
| 2 | 3 |
N must equal 2 and O must equal 3 for the ratio and proportion to be equal.
Example #19: A 150. g sample of a compound is found to be 44.1% C, 8.9% H and the remainder oxygen. What is the compound's empirical formula?
Solution:
1) Since percentages are given, we can assume 100 g (rather than 150 g) of compound is present:
Carbon ---> 44.1 gBecause the percentages are given, the fact that the sample is 150 g in mass is redundant.
2) Convert mass to moles:
C ---> 44.1 / 12.011 = 3.6716
H ---> 8.9/ 1.008 = 8.8294
>O ---> 47.0 / 15.9994 = 2.9376
3) Use the smallest of answers above. Divide it into each answer:
C ---> 3.6716 / 2.9376 = 1.24986 mol = 1.25
H ---> 8.8294 / 2.9376 = 3.0056 = 3.00
O ---> 2.9376 / 2.9376 = 1.00
4) Think about the answers from step 3 as improper fractions:
C ---> 1.25 = 5⁄4
H --->12⁄4
O --->4⁄4
5) Multiply through by 4:
C ---> 5The empirical formula is C5H12O4
6) If your teacher were to insist on you using 150 g, then start this way:
C ---> (150) (0.441) =and then convert the masses to moles and then do the calculations to get to the lowest set of whole-number subscripts.
Example #20: Nitrogen forms more oxides than any other element. The percentage mass of nitrogen in one of the oxides is 36.85%.
(a) Determine the empirical formula of the compound
(b) Determine the molecular formula for this compound, given that its molecular weight is 152.0 g mol¯1
Solution: the typical way . . . .
1) Assume 100 g of compound is present.
2) Convert that %N and 100 g to mass N and mass O
N ---> 36.85 g
O ---> 100 − 36.85 = 63.15 g
3) Convert masses to moles
N ---> 36.85 g / 14.007 g/mol = 2.631 mol
O ---> 63.15 g / 15.9994 g/mol = 3.947 mol
4) Simplify mole ratio to get empirical formula.
N ---> 2.631 mol / 2.631 mol = 1Multiply by 2 to N = 2 and O = 3
N2O3 is the empirical formula
5) Compare molecular mass to empirical unit mass to get number of empirical units per molecule and thus molecular formula.
N2O3 weighs 76.0Quicken 2015 2 3 1. 152.0 / 76.0 = 2
N2O3 times 2 = N4O6
Solution: a different way . . . .
This method depends on knowing the molecular mass. if that value s not provided, we have to use the 'assume 100 g of the compound is present' method.
1) Determine the mass of N and O resent in one mole of the nitrogen oxide:
N ---> (0.3685) (152.0 g) = 56.012 gThe oxygen value could also be arrived at via this:
(0.6315) (152.0 g)
2) Determine moles of each:
N ---> 56.012 g / 14.007 g/mol = 3.9988577I think it's safe to round those answers off to 4 and 6.
3) Write the formulas:
molecular ---> N4O6
empirical ---> N3O3
Bonus Problem: A chemist observed a gas being evolved in a chemical reaction and collected some of it for analyses. It was found to contain 80% carbon and 20% hydrogen. It was also observed that 500. mL of the gas at STP weighed 0.6695 g. What is the empirical formula for the compound? What is its molecular formula?
Solution:
1) Determine the empirical formula:
Assume 100 g of the compound is present.That means 80 g of C and 20 g of H.
That means 6.67 mole of C and 20 mole of H.
The above molar ratio is 1:3, meaning the empirical formula is CH3
2) Determine the molar mass of the compound:
Since everything is at STP, I can use molar volume.| 22.414 L | 0.500 L | |
| ––––––– | = | ––––––– |
| 1.00 mol | x |
x = 0.0223075 mol
molar mass ---> 0.6695 g / 0.0223075 mol = 30.0 g/mol
3) Determine the molecular formula: Curio 13 2 – brainstorming and project management app.
The 'empirical formula weight' (not a standard term in chemistry) of CH330 / 15 = 2
Molecular formula is C2H6.
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