The most recent topic in Chemistry- THE MOLE. The chemical unit is just as ugly as the animal, in my opinion. Although I am still very confused with the subject, I have retained some information about this important unit of measurement. So far, I have learned that the mole is used to express small amounts of matter, such as elements and compounds. This unit provides us with a consistent way to convert between atoms/molecules and grams, and when performing calculations, the mole is a convenient unit.
The mole is defined as “the quantity of a substance that has the same number of particles as are found in 12.000 grams of carbon-12.” This number is equivalent to 6.022×1023, and is called Avogadro’s number. I was surprised by the flexibility of the mole. By this, I mean the way in which it makes any conversion simple. For example, one mole of a compound contains 6.022×10^23 molecules of a compound. The molecular weight of the compound in atomic mass units is the same as the mass in grams of one mole of a compound, making conversions between atoms/molecules, moles, and grams. The formula for determining the number of moles of a sample is below.
weight of sample (g) / molar weight (g/mol)
1. What tasks have you completed recently?
Recently, I have completed coloring packets for anatomy, a worksheet on a Supreme Court case for government, and an empirical equations worksheet for chemistry. I also went to a leadership conference this weekend.
2. What have you learned recently?
Recently, I have learned about the endocrine system, the play “Hamlet”, and the Supreme Court.
3. What are you planning on doing next?
I plan on increasing the duration and intensity of my runs and workouts, finishing my anatomy coloring packets, reading a large chunk of my anatomy glog book, and enjoying a family trip to Denver over spring break.
In chemistry, percent composition is an important tool used to solve problems, especially those relating to empirical formulas. Percent composition, when used correctly, determines what amount of a substance is in a chemical formula. In order to find the percent composition, you need no more than a title and the atomic mass of the elements in the substance. Lets take carbon dioxide (CO2) as an example.
First, one must find the atomic mass of the elements in the formula. The mass (equivalent of one mole) of carbon, is 12.01, and the mass of oxygen is 16.00. Then, those masses must be multiplied by the number of atoms in the compound (CO2), so 12.01×1= 12.01 and 16.00×2= 32.00. After that, the resulting masses (12.01 and 32.00) must be added to find the total mass of the compound, resulting in 44.01. Using the total mass of the compound, the weight of an element in the compound must be divided by the total mass, for example, 12.01g C/44.01 and 32.00g O/44.01, and then multiplied by 100 to find the end result of percent composition. So carbon dioxide (CO2), is 27.29 percent carbon and 72.71 percent oxygen. It is important to use significant figures throughout to make calculations precise, and then if the number needs to be rounded later, it is simple to do.
In our most recent mini lab, we used beans to represent atoms, and from the weight of fifty “atoms”, we the calculated relative mass. After that, we found how many beans were in a pot by adding beans one by one onto the balance until we reached the relative mass. A ‘pot’ of beans is the number of beans it takes to reach the relative mass of the “atom”, or bean.The relative mass is used to determine the number of beans in a pot because the relative mass is an approximation of the mass of the pot.
As we later discovered, the ‘pot’ of beans truly represented a mole. A mole is a unit of measurement that utilizes the relative mass of an atom in finding the amount of any element. This ‘pot’ is a good model for the mole because the mole is based off of the number of atoms in a relative mass, similar to the ‘pot’. When it comes to the mole, the relative mass of hydrogen is used, because it has the smallest atomic weight of all elements.
In our chemistry class, we have recently discussed the importance of accuracy, precision, and significant digits. When using every day terms, accuracy and precision are considered to be synonymous. In the world of science, this couldn’t be less true. Accuracy is defined in the scientific world as the closeness of a measurement of a quantity to that quantity’s actual value. On the other hand, precision is related to the degree in which repeated measurements under unchanged conditions result in the same data. This means that the two are actually quite different, although they may at first appear similar.
The image above displays the differences between accuracy and precision.
Significant digits are a closely related topic. “Sig-Digs”, as they are termed, play very important roles in accuracy and precision. When measuring data in lab, they are crucial to recording precise measurements. Significant digits impact how precise a measurement is, and the number of sig digs present impacts to which place a number is accurate. For example, if you have .5809, the 9 lets you know that the number is accurate to the ten thousandths place. Regarding precision, sig digs are utilized in making sure measurements are “spot on”. If you are measuring a weight and you see that while the scale only measures to the .1, but you can visually see that the weight falls between marks, you can measure to the .01 in order to make the number more precise.
The link above is a video that explains the importance of significant digits in further detail.
In our most recent mini lab, we set out to find the percentage of water in a hydrate. By boiling off the water, we were able to distinguish what weight was CuSO4 and what weight was H2O. We came to the conclusion that 36.8 percent of the hydrate is water. Our results are very reliable; we were careful while performing the experiment and while measuring and massing. That being said, we could have lost some mass that was considered to be water as a result of the substance sticking to the end of the stirring rod. We know that our results are reliable because we followed instructions meticulously and had little to no errors during experimentation. Also, we used significant digits accordingly, therefore improving precision.
Due to our calculations, we found that 2.2 grams of water would be ‘lost’ if you used 6.0g of hydrate. This result is based upon our earlier findings that established the percent composition of water of this sample to be 36.8%. If some of the hydrate/anhydrous solid were splattered out, our results would change, mainly our masses. If some of the hydrate splattered, the percent composition would be inaccurate because the lost mass would be attributed to water, meaning that it would seem more water had been present than actually was. If some of the anhydrous solid splatter out, then the same result would occur, because the lost mass would be seen as water weight.
(90g H2O/249.6g CuSO4 x 5H20) x 100= 36.05
The calculation above shows what we should have gotten as a percent composition for the hydrate in CuSo4 x 5H2O. We had an actual percent composition of 36.8% based off of our experiment.
(2.02g CuSO4 x 5H2O/1) x (90g H2O/ 249.6g CuSO4 x 5H2O)= .728
The equation above displays the calculation of the theoretical mass of hydrate in CuSO4. It was very close to our actual result of .74, meaning our percent error was very low. The equation for percent error can be seen below.
Percent Error (A-T/T) .74-.728/.728 x 100= .016%