Wait a second!
More handpicked essays just for you.
More handpicked essays just for you.
Procedure of chemical reaction lab report
Lab report about chemical reactions
Laboratory reports 4 chemical aspects
Don’t take our word for it - see why 10 million students trust us with their essay needs.
Recommended: Procedure of chemical reaction lab report
Purpose of A Compound
Purpose
i) To determine what happens to the mass when the compound magnesium oxide is formed from the elements magnesium (Mg) and oxygen.
To determine the ratio by mass of the magnesium and the oxygen in the compound magnesium oxide.
Hypothesis
When heated the mass should increase due to the introduction of oxygen to the magnesium. The predicted ratio of magnesium and oxygen will be 3:2 (3 parts magnesium, 2 parts oxygen).
Materials
Equipment Chemicals
Retort Rod Magnesium Ribbon
Retort Base
Ring Clamp
Clay Triangle
Tongs
Bunsen Burner
Crucible
Flint Lighter
Safety Goggles
Procedure
Set up all equipment as demonstrated.
Mass the crucible.
Mass the crucible and the piece of magnesium.
Calculate the
…show more content…
Yes, the possibility of defying a Law of Chemistry was present during out lab. This is believed because the mass of the magnesium was smaller than the compound created, magnesium oxide.
Explain the change in mass that occurred in the formation of magnesium oxide. In the formation of magnesium oxide a change in mass occurs due to the fact that oxygen has been introduced to the magnesium, making magnesium oxide. Since the oxygen and magnesium are chemically bonded it appears to be heavier then the initial weight of the magnesium.
Is the formation of magnesium oxide a physical or chemical change? Why? The formation of magnesium oxide is a chemical change because two substances combined to form a new substance. Magnesium oxide is now a different substance then the magnesium at the beginning of the lab. The formation of chemical bonds allows the magnesium oxide to be a chemical process.
6. Calculate the experimental percent ratio by mass for magnesium and oxygen in magnesium oxide. The experimental ratio by mass for magnesium and oxygen in magnesium oxide would be 3:2.
Mg- 24/40 = 60%
O- 16/40 = 40%
60:40 =
…show more content…
Calculate the actual percent ratio by mass for Mg and O in the compound MgO.
Mg- .115/.14 = 82%
O = 18%
82:18
Calculate the percent error between your experimental and actual percent rations by mass for:
Mg- 82-60 x 100% = 36% 60
O- 18-40 x 100% =
One of the best methods for determining mass in chemistry is gravimetric analysis (Lab Handout). It is essentially using the the mass of the product to figure out the original mass that we are looking for. Thus the purpose of our experiment was to compare the final mass in our reaction to the initial mass and determine the change in mass.
Compared with the accepted value of –601.8 kJ/mol Mg, our experimental error was 2.46%. Introduction In this investigation the change in enthalpy will be determined from the following equation: 2Mg + O2 ® 2MgO, but in an indirect manner. Magnesium metal burns on a bright, extremely hot flame to produce magnesium oxide. It would be difficult to measure the heat of the reaction since the reaction is rapid and occurs at a high temperature (LeMay et al, 1996).
The purpose of this lab was to calculate the percent composition by mass of oxygen in potassium chlorate.
The mass of Mg + the mass of O2=mass of MgxOx. Knowing the mass of
Aim: The aim of this experiment was to determine the empirical formula of magnesium oxide.
In the lab the reaction that took place was a synthesis reaction. A synthesis reaction, is a type of chemical reaction in which two or more simple substances combine to form a new product. The reactants may be elements or compounds. In this case it is a gas and a metal that will react and produce a compound. The general form of a synthesis reaction is, A + B → AB. In order for this lab to be done successful you need knowledge on, percent composition, the empirical and molecular formula, the law of conservation of mass, moles and molar mass, qualitative and quantitative. To begin, the percent composition of a compound is the percent of the total mass that each element has in that compound. Every compound would have a certain percent composition. To calculate percent composition of a compound, you would have to determine the total molecular mass of the compound. For example, for H2O the total molar mass would be 18.00g/mol. You would then input the mass of one of the elements and the molar pass into the equation % by weight (mass) of element = (total mass of element present ÷ total mass of compound) x 100 to find out the percent composition. So for Oxygen it would be, % of O = (16.00g ÷ 18.00g/mol) x 100 which would equal 88.9%. Therefore the percent composition of O in this compound is roughly 88.9%. Furthermore, the molecular formula is the number and types of atoms that are existing in a single molecule of a substance. The empirical formula also known as the simplest formula is the ratio of elements present in the compound. The key difference between these two is that the empirical formula shows the simplest positive integer ratio of atoms of each element present in a compound whereas the molecular formula of a compound is a way ...
== Refer to Chemistry Lab # 2 – Investigating Changes. No changes have been made in this experiment. Results = ==
The primary problem that would occur if magnesium were to be substituted by carbon is that carbon and oxygen form the compound CO2 when they react, which is in a gaseous state above -78.5 degrees Celsius, which is far below the temperature at which the experiment was performed. In this state, it is difficult to contain carbon dioxide gas without an airtight container. This experiment is not possible with an airtight container, since oxygen must enter to react with the carbon, and therefore some the carbon dioxide would inevitably escape, rendering the results invalid.
Other methods available consisted of timing how long it took for the magnesium to disappear. We would place the liquid and the magnesium in a flask and start the timer, one reason we chose not do this is we won’t precisely know when the reaction is finished. Plus people might have different opinions of when the magnesium has disappeared. Another is timing how long it takes for a solution to turn cloudy. We would again place the liquid and the magnesium in a flask and start our timer, the reason...
This would have caused the mass of the product to be much smaller than it was supposed to be. Another scenario that can explain this is that the copper (II) sulfate was heated passed its anhydrous state and caused the breaking down of copper (II) sulfate. In part two of the experiment the formula of the hydrated magnesium sulfate was calculated to be MgSO4 ·8H2O. The formula name would be magnesium sulfate octahydrate. The correct formula is MgSO4·7H2O meaning the name should have been magnesium sulfate heptahydrate. The larger amount of water loss could have been caused by the excess heating of magnesium sulfate resulting in some magnesium sulfate breaking down. Another error could have been that some of the magnesium sulfate was taken out during the stirring process with the spatula. Some magnesium sulfate could have been left on the spatula and not put back in the
Reaction 2: H = 50 x 4.18 x -10.3" H = -2152.7 This value is for 1.37g of calcium oxide, not 56.1g, which is its relative molecular mass. Therefore: H =
The purpose of the lab was to determine the empirical formula of a compound (magnesium oxide). In order to calculate the empirical formula, the mass of each element in the compound was determined. Then, the number of moles of each element in the sample was calculated. Finally, the molar ratio of each element as the smallest whole number was expressed, yielding the compounds empirical formula. The expected empirical formula of Magnesium Oxide was a 1:1 ratio.
Based on your experiments what is the formula of the colorless gas that is released when heating the malachite?
81/125 = 0.648g To calculate how much zinc oxide is produced from 2g of calamine 0.648 is doubled: 0.648 x 2 = 1.296g To calculate how much zinc oxide is produced from 3g of calamine 0.648 is tripled and so on when increasing the mass of calamine. On the next page there is a table of the theoretical conversions for how much zinc oxide is produced from using certain amounts of calamine (1g-9g) Mass Of ZnCO₃ (g) Mass of ZnO Produced (g) 1 0.648 2 1.296 3 1.944 4 2.592 5 3.24 6 3.888 7 4.536 8 5.184 9 5.832 Here is a graph of the results predicted from the table………… This experiment is going to see how much zinc oxide can be obtained from calamine. It shall also show how close to the conversion can be achieved in practice. Hypothesis In a previous experiment where copper carbonate was obtained from malachite the results showed that as the more malachite was used the more product was produced.
3 cm of magnesium ribbon generally has a mass of 0.04 g and yields 40 cm3 of hydrogen when reacted with excess acid. 50 cm3 of 1M hydrochloric in this experiment is in excess.