Explain what is meant by relative atomic mass?-
Relative atomic mass is the measure of the mass of an element which consists of the protons and neutrons; it is always the bigger number and has the symbol Ar. An obvious example of this is oxygen has the Ar of 16 yet sodium has the Ar of 23.
Explain how relative formula mass is calculated using the periodic table-
You need to find the relative atomic mass (M) for each atom but this can only be done when the relative atomic mass have been found, then you add all of the masses in the compound together to find the RFM. Sodium Hydroxide is the example we used within our experiment.
Na= 23
O= 16
H= 1
NaOH 23 + 16 + 1 = 40
NaOH Mr= 40
Explain how it is possible to work out amounts of substance
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Explain how the molarity of the standard solution (the alkali) was calculated in the experiment (equation explained)-
0.1M of NaOH is required, this equation will be used:
Concentration = moles volume
This will be rearranged to find the moles needed to carry out the experiment. The concentration of the experiment using NaOH is 0.1M so we just need to rearrange the equation to find the molarity.
0.1 x 0.250 = 0.0250 moles
Number of moles = mass RFM
0.0250 = mass 40
0.0250 x 40 + 1g (mass)
Explain how this enabled you to accurately calculate the molarity of each acid used in the titrations (equations explained)-
Molarity of the acid = molarity of the alkali x volume of the alkali volume of acid
Firstly we will need to add up all of the volumes found within the titration to find an average:
13.10+13.20+13.10= 13.13
Molarity of Ethanoic acid = 0.1 x 25.00 = 0.190 mol dm-3 13.13
Molarity of Hydrochloric acid = 1.0 x 25.00 = 0.077 mol dm-3 32.53
Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029
To begin this, experiment our group start to weigh three difference empty test tube to get their mass before we put any unknown salt in so we don’t make a calculated mistake. Zeroing the balance with the beaker inside, we put the test tube in the beaker to calculate the unknown hydrate mass.
1mL of Acetic acid was then added to Unknown D and the solution was stirred. Next, 15mL of sodium
In order to begin this experiment, first one must find the balanced chemical equation for the reaction which occurs between the aluminum and copper (II) chloride. This balanced equation being 2Al(s)+3CuCl2 (aq)3Cu(s)+2AlCl3 (aq). After finding this equation, one must use the process of stoichiometry in order to find how many grams of aluminum are needed in order to produce 0.15 grams of copper. In this experiment, the purpose was to produce between 0.1 and 0.2 grams of copper, so one should attempt to produce 0.15 grams of copper seeing as it is the average of those two numbers. The first step in the stoichiometric process which one has to complete is finding how many grams of copper are in one mole of copper.
What is the molecular formula if the molecular weight is 58.0 g/mole? A compound containing Sulfur and Nitrogen is 69.6% Sulfur by mass. If its molar mass is 184 g/mole: What is the empirical formula What is the
In the lab “All That Glitters” the objective that was focused on during the lab was calculating the density, volume and mass of various substances. The method that was used in finding the volume of the samples is called the displacement method. This is a process where the volume of the water in the graduated cylinder is calculated before and after the sample is placed. In this lab, the goal of the experiment was to identify and come to consensus about what the unknown substance might be. For this experiment, the required materials were ten pre and post pennies, unknown sample, graduated cylinder, weigh boat, water, paper towels and a weighing scale.
There is one mole of OH- in the solution since NaOH goes to Na+ and OH-. Trial 1: 25.65mL NaOH x 0.100mol/1000mL = 2.57 x 10-3 mol NaOH = 2.57 x 10-3 mol HA = 2.57 x 10-3 mol H+. The equivalent mass is 0.356g Acid / 2.57 x 10-3 mol H+ = 139g/mol H+ Trial 2: 49.57mL NaOH x 0.100 mol / 1000 mL = 4.96 x 10-3 mol NaOH = 4.96 x 10-3 mol HA = 4.96 x 10-3 mol H+. The equivalent mass is 0.644g Acid / 4.96 x 10-3 mol H+ = 130.g/mol H+ Average =
Using the equation m = ΔTf/Kf , the molality of the unknown solution was found. Then, moles of unknown were calculated, which was used to calculate the average molar mass of unknown. Theory: After the experiment was completed, the data
Procedure: As follows in Mrs. Lubin’s “Finding Moles within a Sample” Lab, found on https://docs.google.com/document/d/1m36zfJCLjGNxoswIMUya_U1HJ4agrHv2yhF6cv1JJU/edit. IV. Data & Calculations: See attached data table. V. Analysis: 1. Define the three identities of the mole.
In this lab there were five different stations. For the first station we had to determine an unknown mass and the percent difference. To find the unknown mass we set up the equation Fleft*dleft = Fright*dright. We then substituted in the values (26.05 N * 41cm = 34cm * x N) and solved for Fright to get (320.5g). To determine the percent difference we used the formula Abs[((Value 1 - Value 2) / average of 1 & 2) * 100], substituted the values (Abs[((320.5 - 315.8) /
A mole is a unit of measurement that follows the rule of ; 1.00 mole = molar mass = 6.02 x 10^23 atoms/ ions/ molecules / formula units = 22.4 L of any gas at standard temperature and pressure. That formula was used to determine the mass of the anhydrated substance. A anhydrate is a substance with water heated out of it. The purpose of this lab was to determine how many moles of water are
Each group was assigned a different percent of sucrose solution out of the four variables; 0% , 5%, 10%, and 15%. After we filled the beaker we then got two potato cores. Once we had the cores we cut the skin off the ends. Following this we then cut the two potato cores into four 2.00 cm potato cores. After they were cut into 2.00 cm each we found the mass.
Procedure A. Preparation of NaOH solution The molarity of a solution is the ratio of the number of solutes dissolved in a liter of solution. To figure out the needed mass (in grams) of NaOH pellets to be dissolved in a 0.25 L of water, remember that a mole is equivalent to the quotient of mass over the molar mass of the substance. This was used to rearrange the base formula and to derive the mathematical equation of mass in terms of molarity. mass (g) =
The mixture were stirred by using a glass rod until the mixture is fully dissolved. The solution were tested by using calibrated pH meter to get the pH value of the solution. Results and Discussions pH ratio between acid and base: 7.3 = 6.82 + x x = 0.48 0.48 = log ([base])/([acid]) 100.48 =base/acid salt/acid = 3.02 There, 1 acid : 3 base calculate number of mole of acid and base to find the mass : molar = mol/L 50 mM = (mol )/(0.5 L) mol = 25 mol number of mole of NaH2PO4 25/4 = acid = 6.25 mol number of mole of Na2HPO4 25/4 × 3 = salt = 18.75 mol to calculate the mass of the acid and base : Mass of NaH2PO4 (6.25 mol)/(119.98 g/mol)=0.052g
The solution turned red when it reached the end point. The titration was continued for 10 seconds after a permanent red color was obtained. The volume of 0.1 M NaOH solution used was determined.