Stoichiometry — Free Chemistry Review Games.

A mole is a counting unit equal to Avogadro's number (6.022 x 10^23), used to count atoms, molecules, or formula units.

Avogadro's number is 6.022 x 10^23, the number of particles in one mole of a substance.

Water has a molar mass of about 18 g/mol (2 hydrogen at 1 g/mol each + 1 oxygen at 16 g/mol).

Coefficients in a balanced equation represent the mole ratios in which reactants combine and products form.

You must first have a balanced equation to determine the correct mole ratios for calculations.

To convert grams to moles, divide the mass in grams by the substance's molar mass (g/mol).

The limiting reagent is the reactant that runs out first, limiting the amount of product that can form.

From the balanced equation \(2H_2 + O_2 \to 2H_2O\), \(2\) moles of \(H_2\) produce \(2\) moles of \(H_2O\).

Percent yield = (actual yield / theoretical yield) x 100%, measuring the efficiency of a reaction.

The excess reagent is the reactant that remains after the reaction is complete because the limiting reagent was consumed first.

\(28\) g \(N_2 = 1\) mol \(N_2\). From the ratio, \(1\) mol \(N_2\) produces \(2\) mol \(NH_3\). \(2\) mol \(\times 17\) g/mol \(= 34\) g \(NH_3\).

Percent yield = (40/50) x 100% = 80%.

10 g H2 = 5 mol; 80 g O2 = 2.5 mol. The ratio needs 2:1 (H2:O2), so 5 mol H2 needs 2.5 mol O2. O2 is exactly consumed, making it the limiting reagent.

Converting to moles: C=3.33, H=6.67, O=3.33. Dividing by smallest (3.33) gives a ratio of 1:2:1, so the empirical formula is CH2O.

Molar ratios from balanced equation coefficients enable conversions between moles of reactants and products, which can then be converted to measurable grams using molar mass.

The molar mass of $NaCl$ is found by adding the atomic masses of sodium (\(22.99\) g/mol) and chlorine (\(35.45\) g/mol), giving \(58.44\) g/mol. The value \(74.55\) g/mol is wrong because that is the molar mass of potassium chloride, $KCl$, not sodium chloride. Always sum the atomic masses from the periodic table according to the subscripts in the formula to find molar mass.

Adding one carbon (\(12.01\) g/mol) and two oxygens (\(2\times16.00=32.00\) g/mol) gives a total of \(44.01\) g/mol for \(CO_2\). The distractor \(28.01\) g/mol is wrong because that is the molar mass of carbon monoxide, \(CO\), which has only one oxygen atom. Molar mass calculations require multiplying each element's atomic mass by its subscript before summing.

Molar mass is defined as the mass in grams of \(6.022\times10^{23}\) particles (one mole) of a substance, linking the atomic scale to a measurable lab quantity. The choice describing atom count is wrong because that describes Avogadro's number's role, not mass. Recognizing molar mass as grams per mole is the key conversion factor for nearly every stoichiometry calculation.

Avogadro's number, \(6.022\times10^{23}\), defines the fixed number of particles—atoms, molecules, or ions—in one mole, regardless of the substance. The option \(6.022\times10^{-23}\) is incorrect because it has the wrong sign on the exponent, making it an impossibly tiny number. This constant is the bridge between the mole and countable particles in every stoichiometric conversion.

The mole is the SI base unit for the amount of substance, representing a specific count of particles just as a dozen represents twelve items. The gram is incorrect because it measures mass, not particle quantity. Distinguishing mass units from the mole is essential before performing any mole-based conversion.

STP is conventionally defined as \(0^\circ C\) (273.15 K) and \(1\) atm of pressure, the conditions under which the molar volume of an ideal gas equals \(22.4\) L. The option \(25^\circ C\) and \(1\) atm describes standard ambient temperature and pressure (SATP), a different reference condition. Knowing the correct STP definition matters because molar volume calculations depend on it.

At STP, one mole of any ideal gas occupies \(22.4\) L, a value derived from the ideal gas law at \(0^\circ C\) and \(1\) atm. The option \(44.8\) L/mol is wrong because that would correspond to two moles of gas, not one. This molar volume lets chemists convert directly between gas volume and moles without needing mass data.

An empirical formula gives the smallest whole-number ratio of atoms present, which may or may not match the true molecular composition. The choice describing the exact atom count is incorrect because that describes a molecular formula, which can be a whole-number multiple of the empirical formula. Students must remember that empirical formulas are ratios, while molecular formulas show actual atom counts.

The molecular formula gives the true, actual number of atoms of each element per molecule, whereas the empirical formula only reduces this to the simplest ratio. The claim that a molecular formula is always identical to the empirical formula is false, since compounds like glucose (\(C_6H_{12}O_6\)) have an empirical formula of \(CH_2O\). This distinction is critical when converting between percent composition data and a compound's true formula.

Mole ratios come directly from the coefficients in a balanced chemical equation, which reflect the fixed proportions in which reactants combine and products form. Molar mass is unrelated to mole ratios, since it only converts moles to grams, not one substance's moles to another's. Every stoichiometric bridge calculation relies on correctly reading these coefficient-based ratios.

Formula mass, calculated in atomic mass units from a chemical formula, becomes molar mass when expressed in grams per mole, since one mole's mass in grams numerically equals the formula mass in amu. The claim that molar mass is always double the formula mass is incorrect and has no chemical basis. This numerical equivalence is what allows chemists to move seamlessly between atomic-scale and lab-scale mass values.

Theoretical yield is the calculated maximum product mass predicted by stoichiometry, assuming the limiting reagent reacts completely with no losses. The option describing product actually collected is wrong because that defines actual yield, a separately measured experimental value. Comparing theoretical and actual yield is the foundation for calculating percent yield.

Actual yield is the experimentally measured mass of product recovered after a reaction, which is typically less than the theoretical prediction due to side reactions or losses during purification. The option describing a stoichiometric prediction is incorrect because that describes theoretical yield, not a measured quantity. Recognizing that actual yield comes from the lab, not calculation, is essential before computing percent yield.

Multiplying moles by molar mass, \(3.5\ \text{mol}\times100.09\ \text{g/mol}=350.3\) g, converts the given amount of substance into a measurable mass. The value \(28.6\) g is incorrect because it results from dividing instead of multiplying, which would be the wrong operation for this conversion direction. Always multiply moles by molar mass when converting from moles to grams.

Since \(18.0\) g of water equals exactly \(1\) mole, multiplying by Avogadro's number gives \(6.022\times10^{23}\) molecules. The value \(3.011\times10^{23}\) is wrong because it represents half a mole's worth of molecules, not a full mole. Converting grams to moles first, then moles to particles, is the standard two-step pathway in mole-particle conversions.

Because the mole ratio of \(H_2\) to \(H_2O\) is \(2:2\), or \(1:1\), \(5\) moles of \(H_2\) produce \(5\) moles of \(H_2O\) when fully reacted. The value \(2.5\) mol is incorrect because it mistakenly applies the \(H_2:O_2\) ratio instead of the \(H_2:H_2O\) ratio. Always match the specific mole ratio between the substances being compared, not just any ratio in the equation.

Fully reacting \(4\) mol \(N_2\) would require \(3\times4=12\) mol \(H_2\), but only \(9\) mol is available, so \(H_2\) runs out first and is limiting. The option naming \(N_2\) is wrong because \(N_2\) is actually present in excess relative to the \(H_2\) supplied. To identify the limiting reagent, always calculate how much of one reactant is needed to consume all of the other, then compare to what is available.

Since the ratio requires \(2\) mol $HCl$ per \(1\) mol \(Zn\), the available \(2.0\) mol $HCl$ can only react with \(1.0\) mol \(Zn\), leaving \(1.0\) mol \(Zn\) unreacted as the excess reagent. The option listing \(1.0\) mol $HCl$ is wrong because $HCl$ is entirely consumed, making it the limiting reagent, not the excess one. Identifying which reagent is limiting first tells you which one is left over and by how much.

Percent yield is calculated as \(\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%=\frac{18.5}{22.0}\times100\%=84.1\%\). The value \(118.9\%\) is wrong because it results from inverting the fraction, which would incorrectly suggest more product than theoretically possible. Percent yield should almost always be less than or equal to \(100\%\) for a pure, correctly calculated product.

The coefficients in the balanced equation directly give the mole ratio of \(Al\) to \(Cl_2\) as \(2:3\). The option \(1:1\) is wrong because it ignores the actual coefficients shown in the equation. Reading mole ratios directly from balanced coefficients, in the correct order, is a foundational stoichiometry skill.

Dividing the given volume by the molar volume at STP, \(\frac{11.2\ \text{L}}{22.4\ \text{L/mol}}=0.5\) mol, gives the number of moles of gas. The option \(22.4\) mol is incorrect because it confuses the molar volume constant itself with the answer to the division. Gas volume at STP can be converted to moles using \(22.4\) L/mol as a direct conversion factor.

Using $moles=Molarity\times Volume(L)$, \(2.0\ \text{M}\times0.250\ \text{L}=0.50\) mol of $NaOH$. The value \(2.0\) mol is wrong because it fails to convert the volume from milliliters to liters before multiplying. Always convert volume to liters before applying the molarity formula to avoid a tenfold or larger error.

The mole ratio of \(Fe\) to \(Fe_2O_3\) is \(4:2\), or \(2:1\), so \(3\) mol \(Fe\) produces \(1.5\) mol \(Fe_2O_3\), which converts to \(1.5\times159.7=239.6\) g. The option \(479.1\) g is wrong because it doubles the correct mass instead of applying the \(2:1\) ratio correctly. Converting moles of one substance to moles of another always requires the correct coefficient ratio before converting to mass.

Percent composition is found by \(\frac{12.01}{44.01}\times100\%=27.3\%\), the fraction of the total molar mass contributed by carbon. The option \(72.7\%\) is wrong because it represents oxygen's percent composition, not carbon's. Percent composition always divides the mass of the element of interest by the total molar mass of the compound.

Converting to moles per \(100\) g, \(\frac{50.0}{32.07}=1.56\) mol \(S\) and \(\frac{50.0}{16.00}=3.13\) mol \(O\), then dividing by the smaller value gives a ratio of \(1:2\), or \(SO_2\). The option \(SO\) is wrong because it does not account for the actual mole ratio obtained from the mass data. Empirical formula problems always require converting mass percentages to moles before finding the simplest ratio.

Dividing the molecular molar mass by the empirical formula mass, \(\frac{180}{30}=6\), gives the multiplier used to scale up the empirical formula subscripts, yielding \(C_6H_{12}O_6\). The option \(C_3H_6O_3\) is wrong because it only applies a multiplier of \(3\), not the correct value of \(6\). Always find this whole-number multiplier before writing the true molecular formula.

Fully reacting \(4\) mol \(Na\) would require \(2\) mol \(Cl_2\), but only \(1.5\) mol is available, so \(Cl_2\) is consumed first and limits the reaction. The option stating both react completely is wrong because \(Na\) is left over once all the \(Cl_2\) is used up. Comparing the required amount of one reactant against what's available is the standard method for identifying the limiting reagent.

Since \(1.5\) mol \(Cl_2\) reacts with \(2\times1.5=3\) mol \(Na\), the remaining unreacted \(Na\) is \(4-3=1\) mol. The option \(2\) mol is wrong because it fails to correctly apply the \(2:1\) mole ratio between \(Na\) and \(Cl_2\). After identifying the limiting reagent, always calculate the exact amount of excess reagent consumed to find what remains.

With a \(1:1\) mole ratio, \(0.25\) mol of limiting reagent forms \(0.25\) mol of product, which converts to mass as \(0.25\times106=26.5\) g. The option \(106\) g is wrong because it ignores the actual number of moles reacting and just restates the molar mass. Theoretical yield calculations always require converting moles of limiting reagent to moles of product before applying molar mass.

Dividing mass by molar mass, \(\frac{75.0}{158.04}=0.475\) mol, converts the given mass into moles. The option \(2.11\) mol is wrong because it results from inverting the calculation, dividing molar mass by mass instead of the reverse. Grams-to-moles conversions always divide the given mass by the molar mass, not the other way around.

Each formula unit of \(Al_2(SO_4)_3\) contains \(2\) aluminum, \(3\) sulfur, and \(12\) oxygen atoms, totaling \(17\) atoms, so \(2.0\) moles contain \(2.0\times17=34\) moles of atoms. The option \(17\) mol is wrong because it only accounts for one mole of the compound, not the two moles given. Always multiply the atoms per formula unit by the number of moles of compound to find total moles of atoms.

Since \(5.0\) mol $CaCO_3$ requires \(2\times5.0=10.0\) mol $HCl$ to react completely, the excess $HCl$ remaining is \(12.0-10.0=2.0\) mol. The option \(10.0\) mol is wrong because that is the amount of $HCl$ consumed, not the amount left over. Determining leftover excess reagent always requires subtracting the amount consumed from the total amount initially provided.

Converting to moles gives \(5.0\) mol \(H_2\) and \(1.786\) mol \(N_2\); since \(1.786\) mol \(N_2\) would require \(5.36\) mol \(H_2\) but only \(5.0\) mol is available, \(H_2\) is limiting, so moles of \(NH_3=5.0\times\frac{2}{3}=3.33\) mol, giving a mass of \(3.33\times17.0=56.7\) g. The option \(85.0\) g is wrong because it assumes \(N_2\) is limiting and overestimates the product formed beyond what the available \(H_2\) can support. Multi-step limiting reagent problems always require converting both reactants to moles and comparing required versus available amounts before calculating product mass.

A percent yield above \(100\%\) is chemically impossible for a pure product, so it almost always signals contamination, such as leftover solvent adding extra mass to the measured sample. The option claiming the law of conservation of mass was violated is wrong because mass is always conserved; the discrepancy comes from measurement error, not a physical law breaking down. Whenever percent yield exceeds \(100\%\), students should suspect impurities or measurement errors rather than an actual surplus of product.

Moles of carbon equal moles of \(CO_2\), \(\frac{8.80}{44.01}=0.200\) mol, and moles of hydrogen equal twice the moles of \(H_2O\), \(2\times\frac{3.60}{18.02}=0.400\) mol, giving a \(C:H\) ratio of \(1:2\), or \(CH_2\). The option \(CH_4\) is wrong because it would require twice as much hydrogen relative to carbon than the combustion data actually shows. Combustion analysis always converts \(CO_2\) and \(H_2O\) masses back into moles of carbon and hydrogen before finding the simplest ratio.

Fully reacting \(6.0\) mol \(Al\) would require \(3\times\frac{6.0}{2}=9.0\) mol \(Br_2\), but only \(8.0\) mol is available, so \(Br_2\) runs out first and limits the reaction. The inert diluent gas is wrong because it does not participate in the reaction at all and therefore cannot be a limiting reagent. Only substances that actually appear in the balanced equation as reactants can ever be considered limiting or excess reagents.

Percent yield relative to the adjusted theoretical yield is \(\frac{36.0}{40.5}\times100\%=88.9\%\), using the corrected baseline rather than the original uncorrected value. The option \(80.0\%\) is wrong because it incorrectly compares the actual yield to the original \(45.0\) g theoretical yield instead of the side-reaction-adjusted value. When side reactions consume reagent, percent yield must be calculated against the realistic adjusted theoretical maximum, not the idealized one.

Converting to moles gives \(4.0\) mol \(H_2\) and \(1.0\) mol \(O_2\); since \(1.0\) mol \(O_2\) only needs \(2.0\) mol \(H_2\), \(O_2\) is limiting and \(2.0\) mol \(H_2\) (or \(4.0\) g) remains unreacted. The option \(16.0\) g \(O_2\) is wrong because \(O_2\) is fully consumed as the limiting reagent, leaving none behind. Determining excess reagent mass always requires first confirming which reactant is limiting, then subtracting the amount of the other reactant actually consumed.

Rearranging the percent yield formula gives theoretical yield \(=\frac{\text{actual yield}}{\text{percent yield}}=\frac{75.0}{0.82}=91.5\) g, the target amount needed before accounting for typical losses. The option \(61.5\) g is wrong because it results from multiplying rather than dividing by the yield fraction, which underestimates the required theoretical amount. When working backward from a desired actual yield, always divide by the decimal form of the percent yield rather than multiply.

Moles present are \(0.500\times0.400=0.200\) mol $AgNO_3$ and \(0.300\times0.500=0.150\) mol $NaCl$; since the ratio is \(1:1\), the smaller amount, $NaCl$, is limiting and produces \(0.150\) mol $AgCl$. The option stating $AgNO_3$ is limiting is wrong because $AgNO_3$ is present in greater molar quantity and is actually the excess reagent. Solution stoichiometry problems require first converting molarity and volume into moles before comparing reactant quantities using the balanced equation's ratio.

Since \(\frac{88.11}{44.05}=2\), the empirical formula subscripts double to give the molecular formula \(C_4H_8O_2\), and \(\frac{17.6}{88.11}=0.200\) mol is present in the sample. The option listing \(0.400\) mol with \(C_4H_8O_2\) is wrong because it uses the empirical formula mass instead of the correct molecular molar mass when converting grams to moles. Once the molecular formula and its true molar mass are established, all subsequent mole conversions must use that corrected molar mass.

Converting gas volume to moles gives \(\frac{44.8}{22.4}=2.0\) mol \(H_2\); since \(2.0\) mol \(N_2\) would require \(6.0\) mol \(H_2\) but only \(2.0\) mol is available, \(H_2\) is limiting, producing \(2.0\times\frac{2}{3}=1.33\) mol \(NH_3\). The option claiming \(N_2\) is limiting is wrong because \(N_2\) is actually present in relative excess compared to the available \(H_2\). Problems that mix gas volumes and mole quantities always require converting the gas volume to moles first using the molar volume at STP before comparing reactant amounts.

Using the mole ratios, \(2.5\) mol \(C_3H_8\) produces \(3\times2.5=7.5\) mol \(CO_2\) and \(4\times2.5=10.0\) mol \(H_2O\), giving a combined total of \(7.5+10.0=17.5\) mol of gaseous products. The option \(12.5\) mol is wrong because it only sums part of one product's moles rather than correctly totaling both \(CO_2\) and \(H_2O\) separately. When a reaction produces multiple products, each must be calculated individually from its own coefficient ratio before combining totals.

The stoichiometric requirement of \(B\) is \(2.0\times3.0=6.0\) mol, and supplying \(20\%\) excess means multiplying by \(1.20\), giving \(6.0\times1.20=7.2\) mol of \(B\) actually used. The option \(6.0\) mol is wrong because it represents only the exact stoichiometric amount, ignoring the specified \(20\%\) excess that was actually supplied. Percent excess problems always require calculating the exact stoichiometric amount first, then scaling up by the given excess percentage.

The intended \(30\%\) excess requires \(4.0\times1.30=5.2\) mol of \(X\), but only \(4.8\) mol is available, meaning \(X\) falls short of the planned excess and actually becomes the limiting reagent instead of \(Y\). The option claiming \(4.8\) mol exactly matches the intended excess is wrong because \(4.8\) mol is less than the calculated \(5.2\) mol target, not equal to it. This scenario shows that even a reagent intended to be in excess can unexpectedly become limiting if the supplied amount falls short of the calculated requirement.