Recently, i have completed the review for the sba and the calculations for our mole lab.

Recently, i have learned the concept of the mole and its role in chemistry.

Next, i plan on furthering my knowledge of the mole and how to use it in calculation

# Monthly Archives: March 2014

# beans in a pot

using beans, we learned more about relative masses and “pots.” A “pot” of beans is simply the amount of beans that make up the relative mass of an “element” in a group of “elements”. the relative mass is used because it shows the amount of pots in relation to the relative mass of the other beans. The pot is a model of a mole in the fact that it is a unit of amount of beans in a certain mass of beans, and a mole is a unit of amount of atoms in a certain amount of anything. they’re both simply units of quantity. the relative mass of carbon is used, because it is the most exact, one mol of carbon atoms is 12g of carbon 12.

# 3 questions for 3/7/14

1. recently i have oxidized magnesium in order to create magnesium oxide.

2. i have learned recently how to solve for empirical formulas

3. i hope to achieve a greater grasp of math in science.

# how to: percent composition.

to find the percent composition of a certain component of a compound, one needs only to find the molar mass of said particular part of a compound and divide it by the total mass of the compound. for example; to find the percent composition of CO2, you need to find the molar mass of carbon, (12.01) and the molar mass of oxygen, (16.00). multiply the masses by number of atoms in the compound, (16.00×2) (12.01×1) and you can find the total mass (44.02) then divide the totla mass of oxygen in the compound by that number and you can find the percentage. (32/44.01) the percentage of oxygen in CO2 is.7271, or rounded and put into percentage, 73%. empirical formulas are just the opposites of percent composition, it gives the percentages as givens, from there, you use molar masses to find the compound. you simply divide the mass of the component and divide it by the molar mass of the same component, from there, divide each of the products by the smallest product. This will give you the number of certain atoms in the compound. From there you can determine the empirical formula.

# science by numbers

to be accurate, is to be close to the actual value of a given piece of information. and to be precise is to have multiple repeatable measurements in range. these attributes are reinforced by the use of significant digits, allowing calculations to maintain similar precision throughout the process without integrating false variables in the equation. these numbers allow a greater understanding of the workings of chemistry. in the article, it says that several elements had their masses changed to be more accurate. these more accurate numbers are a part of science, the constant search for more accurate information through more precise calculation (in this case)

# Hydrate composition

The hydrate we used was copper (II) sulfate pentahydrate, the water percentage is 37%, our results are fairly reliable, using the periodic table, we were able to determine through mass the percentage on an atomic scale. the percentage determined through this method predicted a percentage of 36.05%, so our percentage was fairly accurate. Our percentage applied to a 6.0g sample would yield a loss of 2.23g of water. if we were attempting to determine the percentage and some spilled out, we would have a higher percentage of perceived water composition due to thee calculation for percentage being the original weight of the substance minus the post-heated sample over 100. If some hydrate spilled, the change in mass would be greater, and the water would be calculated to be a higher percentage.

[update]

the theoretical percentage can be determined by the equation [(90gH20)/(249.6gCuSO4~5H2O)] X100, producing an ideal percentage of 36.05

the theoretical amount of water in our sample can be determined by the equation 2.29g CuSO4~5H2O/1 X 90.0g H2O/249.6g CuSO4~5H2O. this gives a theoretical water weight of .83g.

our recovered water was .85g, so our percent error was .85-.83=.02×100=2% error