Stoichiometry is the accounting, or math, behind chemistry. Given enough information, one have the right to use stoichiometry to calculation masses, moles, and also percents in ~ a chemistry equation.

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What is a chemistry EquationThe MoleBalancing chemical EquationsLimiting ReagentsPercent CompositionEmpirical and Molecular FormulasDensityConcentrations that Solutions

What is a chemistry equation?

In chemistry, us use symbols to stand for the various chemicals. Successin chemistry relies upon arising a strong familiarity v these basic symbols. For example, the price "C"represents an atom of carbon, and also "H" to represent an atom the hydrogen. To stand for a molecule of table salt, salt chloride, we would use the notation "NaCl", whereby "Na" represents sodium and also "Cl" represents chlorine. We call chlorine "chloride" in this case because of its connection to sodium. You will have a chance to review naming schemes, or nomenclature
, in a later reading.A chemical equation is one expression that a chemistry process. Because that example:AgNO3(aq) + NaCl(aq) ---> AgCl (s) + NaNO3(aq)In this equation, AgNO3 is combined with NaCl. The equation mirrors that the reaction (AgNO3 and NaCl) react v some process (--->) to kind the assets (AgCl and also NaNO3). Since they experience a chemical process, castle are readjusted fundamentally. Regularly chemical equations are written mirroring the state the each problem is in. The (s) sign method that the compound is a solid. The (l) sign method the substance is a liquid. The (aq) authorize stands because that aqueous in water and way the link is liquified in water. Finally, the (g) sign method that the compound is a gas. Coefficients are provided in every chemical equations to display the relative amounts of each substance present. This amount deserve to represent one of two people the relative number of molecules, or the relative number of moles (described below). If no coefficient is shown, a one (1) is assumed. On part occasions, a variety of information will it is in written over or below the arrows. This information, such as a value for temperature, display what conditions need come be existing for a reaction come occur. For example, in the graphics below, the notation above and below the arrows mirrors that we need a chemical Fe2O3, a temperature the 1000 degrees C, and also a pressure of 500 environments for this reaction to occur.The graphic below works to record most the the principles described above:
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The Mole

Given the equation above, we deserve to tell the variety of moles the reactants and also products. A mole merely represents Avogadro"s number (6.023 x 1023) of molecules. A mole is similar to a term like a dozen. If you have a dozen carrots, you have twelve the them. Similarily, if you have actually a mole the carrots, you have 6.023 x 1023 carrots. In the equation above there room no numbers in prior of the terms, so every coefficient is suspect to it is in one (1). Thus, you have the same variety of moles of AgNO3, NaCl, AgCl, NaNO3.Converting between moles and grams that a problem is often important.This conversion deserve to be easily done when the atomic and/or molecularweights that the substance(s) room known. The atom or molecularweight of a substance in grams provides up one mole that the substance.For example, calcium has an atomic load of 40 grams. So, 40grams of calcium makes one mole, 80 grams renders two moles, etc.

Balancing chemistry Equations

Sometimes, however, we have to do part work before using the coefficients that the terms to represent the relative variety of molecules of every compound. This is the instance when the equations space not appropriately balanced. We will take into consideration the following equation:Al + Fe3O4---> Al2O3Since no coefficients room in front of any kind of of the terms, it is straightforward to assume that one (1) mole the Al and also one (1) mole that Fe304 react to kind one (1) mole the Al203. If this were the case, the reaction would be rather spectacular: an aluminum atom would show up out that nowhere, and two (2) steel atoms and also one (1) oxygen atom would certainly magically disappear. We understand from the legislation of conservation of fixed (which says that matter can neither be produced nor destroyed) that this simply cannot occur. We need to make certain that the number of atoms the each certain element in the reactants equals the variety of atoms of the same facet in the products. To execute this we have actually to number out the relative number of molecules of each term to express by the term"s coefficient.Balancing a chemistry equation is basically done through trial and error. Over there are many different ways and systems of act this, however for every methods, it is important to know exactly how to count the number of atoms in an equation. For example we will certainly look in ~ the following term.2Fe3O4This ax expresses two (2) molecules of Fe3O4. In every molecule of this substance there space three (3) Fe atoms. Thus in 2 (2) molecule of the substance there must be 6 (6) Fe atoms. An in similar way there are 4 (4) oxygen atoms in one (1) molecule the the problem so there have to be eight (8) oxygen atom in 2 (2) molecules. Now let"s shot balancing the equation discussed earlier:Al + Fe3O4---> Al2O3+ Fe occurring a strategy have the right to be difficult, however here is one way of pull close a difficulty like this. count the number of each atom ~ above the reactant and also on the product side. Determine a term come balance first. When looking in ~ this problem it shows up that the oxygen will be the most challenging to balance so we"ll try to balance the oxygen first. The simplist way to balance the oxygen state is:Al +3 Fe3O4---> 4Al2O3+Fe that is necessary that you never change a subscript. Only readjust the coefficient once balancing one equation. Also, be sure to an alert that the subscript times the coefficient will offer the number of atoms of that element. Top top the reactant side, we have actually a coefficient of 3 (3) multiplied by a subscript of 4 (4), offering 12 oxygen atoms. Top top the product side, we have actually a coefficient of four (4) multiply by a subscript of three (3), providing 12 oxygen atoms. Now, the oxygens are balanced. Choose an additional term to balance. We"ll choose iron, Fe. Since there space nine (9) iron atoms in the hatchet in i beg your pardon the oxygen is balanced we include a ripe (9) coefficient in former of the Fe. We currently have:Al +3 Fe3O4---> 4Al2O3+9Fe Balance the last term. In this case, because we had actually eight (8) aluminum atoms on the product next we require to have actually eight (8) top top the reactant next so we include an eight (8) in former of the Al hatchet on the reactant side. Now, we"re done, and also the balanced equation is:8Al + 3Fe3O4 ---> 4Al2O3 + 9 Fe

Limiting Reagents

Sometimes once reactions occur between two or an ext substances, onereactant runs out before the other. The is dubbed the "limitingreagent." Often, it is important to recognize the limiting reagent in a problem. Example: A chemist only has actually 6.0 grams the C2H2 and also an unlimitted supply of oxygen and also desires to develop as lot CO2 together possible. If she supplies the equation below, how much oxygen should she include to the reaction?2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2 H2O(l) To resolve this problem, that is important to determine exactly how much oxygen shouldbe added if all of the reactants were supplied up (this is the means to create the maximum lot of CO2). First, us calculate the variety of moles of C2H2 in 6.0 grams of C2H2. To be able to calculate the moles we need to look at a routine table and also see that 1 mole that C weighs 12.0 grams and H weighs 1.0 gram. Thus we understand that 1 mole of C2H2 weighs 26 grams (2*12 grams + 2*1 gram). Due to the fact that we only have 6.0 grams that C2H2 we must discover out what portion of a mole 6.0 grams is. To do this, we use the adhering to equation.
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Then, because there are 5 (5) molecules of oxygen come every two (2) molecule of C2H2, we must multiply the moles of C2H2 by 5/2 to gain the full moles the oxygen that would be offered to react with all the C2H2. Us then convert the moles of oxygen to grams in order to find the amount of oxygen that requirements to be added:
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Percent Composition

It is feasible to calculate the mole ratios (also referred to as mole fractions) in between terms in a chemistry equation when given the percent by fixed of products or reactants. Percentage by mass = massive of part/ mass of wholeThere are two types of percent ingredient problems-- problems in which you are given the formula (or the load of every part) and asked to calculation the percent of every elementand problems in which you are offered the percentages and also asked to calculation the formula.In percent ingredient problems, there are many possible solutions. The is always possible to twin the answer. For example, CH and also C2H2 have the same proportions, yet they are various compounds. That is standard to provide compounds in their easiest form, whereby the ratio between the facets is asreduced as it deserve to be-- referred to as the empirical formula. When calculating the empirical formula from percent composition, one can transform the percentages come grams. Because that example, the is normally the simplest to i think you have actually 100 grams therefore 54.3% would become 54.3 grams. Then we can transform the masses to moles which offers us mole ratios. It is vital to alleviate to totality numbers. A an excellent technique is to divide all the state by the smallest variety of moles. Then the ratio of the moles have the right to be transfered to write the empirical formula.Example: If a compound is 47.3% C (carbon), 10.6% H (hydrogen) and 42.0% S (sulfur), what is that empirical formula? To perform this difficulty we must transfer all of our percents to masses. Us assume the we have actually 100 g that this substance. Climate we convert to moles:
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Now we try to gain an also ratio in between the aspects so we divide by the variety of moles that sulfur, because it is the the smallest number:
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So us have: C3H8 SExample: figure out the percentage by massive of hydrogen sulfate, H2SO4.In this problem we need to an initial calculate the complete weight the the compound by looking in ~ the routine table. This gives us:(2(1.008) + 32.07 + 4(16.00) grams/mol = 98.09 g/mol Now, we need to take the weight portion of each aspect over the full mass (which we just found) and also multiply by 100 to obtain a percentage.
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Now, us can inspect that the percentages add up to 100% 65.2 + 2.06 + 32.7 = 99.96This is essentially 100 so we recognize that everything has worked, and we probably have actually not made any kind of careless errors. so the prize is the H2SO4 is consisted of of 2.06% H, 32.7% S, and also 65.2% O through mass.

Empirical Formula and Molecular Formula

While the empirical formula is the simplest type of a compound, themolecular formula is the form of the term as it would appear in a chemicalequation. The empirical formula and also the molecule formula deserve to be thesame, or the molecular formula have the right to be any multiple that the empiricalformula.Examples that empirical formulas: AgBr, Na2S, C6H10O5. Examples of molecular formulas: P2, C2O4, C6H14S2, H2, C3H9.One have the right to calculate the empirical formula indigenous the masses or percentage composition of any type of compound. We have currently discussed percent ingredient in the section above. If we only have mass, every we space doing is essentially eliminating the step of convertingfrom percentage to mass. Example: calculate the empirical formula because that a compound that has 43.7 g ns (phosphorus) and also 56.3 grams of oxygen.First we convert to moles:
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Next we divide the moles to shot to gain a even ratio.
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When we divide, we did not get entirety numbers for this reason we have to multiply by 2 (2). The answer=P2O5Calculating the molecular formula once we have actually the empirical formula is easy. If we know the empirical formula the a compound, all we should do is divide the molecular mass the the link by the fixed of the empirical formula.It is also feasible to carry out this with among the facets in the formula;simply division the fixed of that facet in one mole of link by the massof that element in the empirical formula. The result should constantly be awhole number. Example: if we recognize that the empirical formula of a compound is HCN and also we room told the a 2.016 grams that hydrogen room necesary to do the compound, what is the molecule formula? In the empirical formula hydrogen weighs 1.008 grams. Splitting 2.016 by 1.008 we watch that the quantity of hydrogen essential is double as much. Because of this the empirical formula needs to be increased by a element of 2 (2). The answer is: H2C2N2.

Density

Densityrefers to the mass every unit volume the a substance. That is a very commonterm in chemistry.

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Concentrations of Solutions

The concentration that a equipment is the "strength" that a solution. A solution frequently refers to the dissolve of part solid problem in a liquid, such as dissolving salt in water. That is additionally often essential to number out exactly how much water to include to a solution to adjust it to a details concentration. The concentration the a solution is typically given in molarity.Molarity is defined as the variety of moles of solute (what is actually liquified in the solution) divided by the liters of solution (the complete volume of what is dissolved and also what it has been dissolved in).
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Molarity is probably the most frequently used term since measuring a volume of fluid is a reasonably easy point to do.Example: If 5.00 grams the NaOH are dissolved in 5000 mL that water, what is the molarity that the solution?One the our first steps is to convert the quantity of NaOH offered in grams right into moles:
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Now we merely use the definition of molarity: moles/liters to get the answer
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So the molarity (M) of the equipment is 0.025 mol/L.Molality is one more common measure up of concentration. Molality is characterized as mole of solute separated by kilograms the solvent (the substance in which it is dissolved, prefer water).
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Molality is periodically used in location of molarity at excessive temperatures since the volume can contract or expand. Example: If the molality the a equipment of C2H5OH dissolved in water is 1.5 and the load of the water is 11.7 kg, figure out how much C2H5OH must have actually been added in grams to the solution? Our first step is to instead of what us know right into the equation. Climate we shot to deal with for what us don"t know: moles of solute. As soon as we recognize the moles of solute we have the right to look at the regular table and also figure out the switch from mole to grams.
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It is possible to convert between molarity and also molality. The only info needed is density.Example: If the molarity the a equipment is 0.30 M, calculate the molalityof the solution understanding that the thickness is 3.25 g/mL.To do this problem we deserve to assume one (1) liter of solution to do thenumbers easier. We require to acquire from the molarity systems of mols/Liter tothe molality units of mols/kg. We occupational the difficulty as follows,remembering the there space 1000 mL in a Liter and also 1000 grams in a kg. Thisconversion will only be specific at tiny molarities and molalities.
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It is also feasible to calculation colligative properties, such together boiling suggest depression, making use of molality. The equation because that temperature depression or growth is change in T= K * mWhere:T is temperature depression (for freeze point) or temperature growth (for cook point) (°C)K is the freeze point constant (kg °C/moles)m is molality in moles/kgExample: If the freezing allude of the salt water put on roads is -5.2 C, what is the molality the the solution? (The Kf for water is 1.86 C/m.) This is a an easy problem whereby we just plug in numbers right into the equation. One item of details we do need to know is the water normally freezes in ~ 00C. T=K * m T/K= mm = 5.2/1.86m = 2.8 mols/kg

Practice Problems

1. If only 0.25 molar NaOH and also water are available, just how much NaOH requirements to be included to do 10 liters that 0.2 molar systems of NaOH? check your work2. If 2.0 moles of sucrose weighing 684 grams is put in 1000 grams that water and is climate dissolved, what would certainly be the molality the the solution?Check your work. 3. If you have actually a 0.25 molar solution of benzene v a density of 15 grams/liter, calculation the molality that the solution. Examine your work4. If the density of mercury is 13.534 g/cm2 and also you have 62.5 cm3 the mercury, how numerous grams, moles, and also atoms the mercury carry out you have? (Mercury has a mass of 200.6 g/mol.) check your work developed by
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