[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-1-en-105":3,"doc-seo-572000-105":53,"doc-detail-572000-en":127},{"code":4,"msg":5,"data":6},0,"success",[7,14,19,24,29,34,39,44,49],{"id":8,"doc_module":9,"doc_module_name":10,"category_name":11,"show_sort_weight":12,"slug":13},11,1,"Template","Presentations",90,"presentations",{"id":15,"doc_module":9,"doc_module_name":10,"category_name":16,"show_sort_weight":17,"slug":18},12,"Resumes",80,"resumes",{"id":20,"doc_module":9,"doc_module_name":10,"category_name":21,"show_sort_weight":22,"slug":23},14,"Invoices",70,"invoices",{"id":25,"doc_module":9,"doc_module_name":10,"category_name":26,"show_sort_weight":27,"slug":28},15,"Posters",60,"posters",{"id":30,"doc_module":9,"doc_module_name":10,"category_name":31,"show_sort_weight":32,"slug":33},16,"Social Media",50,"social-media",{"id":35,"doc_module":9,"doc_module_name":10,"category_name":36,"show_sort_weight":37,"slug":38},17,"Forms",40,"forms",{"id":40,"doc_module":9,"doc_module_name":10,"category_name":41,"show_sort_weight":42,"slug":43},18,"Letters",30,"letters",{"id":45,"doc_module":9,"doc_module_name":10,"category_name":46,"show_sort_weight":47,"slug":48},21,"Paper Templates",5,"papers-templates",{"id":50,"doc_module":9,"doc_module_name":10,"category_name":51,"show_sort_weight":4,"slug":52},158,"General","general-158",{"code":4,"msg":54,"data":55},"ok",{"site_id":56,"language":57,"slug":58,"title":59,"keywords":60,"description":61,"schema_data":62,"social_meta":120,"head_meta":122,"extra_data":124,"updated_unix":126},105,"en","moles-stoichiometry-cheat-sheet","Moles & Stoichiometry Cheat Sheet","","Moles & Stoichiometry Cheat Sheet provides step-by-step methods for key chemistry calculations, including molar mass, percent composition, empirical and molecular formulas, and hydrate formulas and naming. It also summarizes core stoichiometry workflows using balanced chemical equations, covering mole-to-mole, mole-to-mass, mass-to-mass conversions, limiting and excess reactants, and percent yield. Each section translates problem data into moles, applies mole ratios, and converts to final requested quantities.",{"@graph":63,"@context":119},[64,80,102],{"@type":65,"itemListElement":66},"BreadcrumbList",[67,71,74,77],{"item":68,"name":69,"@type":70,"position":9},"https://docshare.wps.com","Home","ListItem",{"item":72,"name":10,"@type":70,"position":73},"https://docshare.wps.com/template/",2,{"item":75,"name":51,"@type":70,"position":76},"https://docshare.wps.com/template/general/",3,{"item":78,"name":59,"@type":70,"position":79},"https://docshare.wps.com/template/moles-stoichiometry-cheat-sheet/572000/",4,{"url":78,"name":59,"@type":81,"image":82,"author":87,"headline":59,"publisher":90,"fileFormat":93,"inLanguage":57,"description":61,"dateModified":94,"datePublished":95,"encodingFormat":93,"isAccessibleForFree":96,"interactionStatistic":97},"DigitalDocument",{"url":83,"@type":84,"width":85,"height":86},"https://docshare.wps.com/thumbnails/moles-stoichiometry-cheat-sheet/572000.png","ImageObject",442,249,{"name":88,"@type":89},"Eliana","Person",{"url":68,"name":91,"@type":92},"DocShare","Organization","application/pdf","2026-10-23","2026-10-13",true,{"@type":98,"interactionType":99,"userInteractionCount":101},"InteractionCounter",{"@type":100},"ViewAction",6,{"@type":103,"mainEntity":104},"FAQPage",[105,111,115],{"name":106,"@type":107,"acceptedAnswer":108},"How do you calculate molar mass of a compound?","Question",{"text":109,"@type":110},"Write the compound’s formula, determine the number of atoms of each element, multiply by each element’s atomic mass, then add all element contributions.","Answer",{"name":112,"@type":107,"acceptedAnswer":113},"What is the process for finding an empirical formula?",{"text":114,"@type":110},"Convert percent composition to grams (use 100 g if needed), convert grams to moles, divide by the smallest mole value to get the simplest ratio, then multiply to remove fractions and assign the element symbols.",{"name":116,"@type":107,"acceptedAnswer":117},"How do you determine the limiting reactant and percent yield?",{"text":118,"@type":110},"Convert reactant amounts to moles and compare moles divided by each reactant’s balanced-equation coefficient; the smaller value indicates the limiting reactant. For percent yield, compute theoretical yield using stoichiometry, divide actual yield by theoretical yield, then multiply by 100.","https://schema.org",{"og:url":78,"og:type":121,"og:title":59,"og:site_name":91,"og:description":61},"article",{"robots":123,"canonical":78},"index,follow",{"doc_id":125,"site_id":56},572000,1792097022,{"code":4,"msg":5,"data":128},{"doc_id":125,"user_id":129,"nickname":88,"user_avatar":130,"doc_module":9,"category_id":50,"category_name":51,"doc_title":59,"doc_description":61,"doc_content":131,"file_id":132,"file_url":133,"file_type":134,"file_size":135,"view_count":101,"is_deleted":4,"is_public":9,"is_downloadable":9,"audit_status":9,"page_count":73,"language":136,"language_code":57,"site_id":56,"html_lang":57,"table_of_contents":137,"faqs":138,"seo_title":139,"seo_description":61,"update_tm":140,"read_time":9},4398048949847,"https://ap-avatar.wpscdn.com/avatar/400002536579ef2da7f?_k=1778318612642679267","Moles & Stoichiometry Cheat Sheet  \n\n| Calculating Molar Mass\u003Cbr>1. Write out formula of compound\u003Cbr>2. Determine number of atoms of each element present\u003Cbr>3. multiply number of atoms of the element by the atomic mass of the element\u003Cbr>4. add up the values just determined for all of the elements present in the compound |  | Calculating Percent Composition\u003Cbr>1. Calculate the molar mass of the compound\u003Cbr>2. divide the mass contribution of each element by the total molar mass\u003Cbr>3. multiply by 100 to make a percent\u003Cbr>Note: total of percents should equal 100%! |  |  |\n| --- | --- | --- | --- | --- |\n| Mass (grams) | x by molar mass\u003Cbr>\u003C   Mole\u003Cbr>  >\u003Cbr>÷ by molar mass |  | x by Avogadro’s \\#   >\u003C ÷ by Avogadro’s \\# | Particles\u003Cbr>(atoms,\u003Cbr>molecules,\u003Cbr>formula units) |\n| Calculating Empirical Formula\u003Cbr>1. Change % of each element to grams of each element (if no sample amount is given assume 100 g)\u003Cbr>2. convert grams of each element to moles of each element by dividing the grams by the molar (atomic) mass of that element\u003Cbr>3. divide all molar values by the smallest molar value (to establish lowest whole number ratio)\u003Cbr>4. if all values are not whole numbers, multiply all by a whole number factor to make the values whole (ex. 1.5 and 1 should be multiplied by 2 to get rid of the .5 = 3 and 2)\u003Cbr>5. plug values in with the element they represent |  | Calculating Molecular Formula\u003Cbr>1. calculate the empirical formula first\u003Cbr>2. calculate the molar mass of the empirical formula\u003Cbr>3. take the molar mass of the molecular formula (given in the problem) and divide it by the molar mass of the empirical formula. This will equal the number of times the [e.f. is](e.f. is) in the m.f.\u003Cbr>4. distribute the number attained through the empirical formula\u003Cbr>Example: 2(C2 H5 ) = C4 H10 |  |  |\n| Calculating the Formula of a Hydrate\u003Cbr>1. Determine the mass of water in the hydrate by subtracting the mass of the anhydrous from the mass of the hydrate (if necessary) -OR-convert %’s into grams by assuming a 100g sample\u003Cbr>2. convert grams of each into moles of each by dividing by molar mass\u003Cbr>3. divide the moles of water by the moles of the anhydrous to determine the number of moles of water in the formula\u003Cbr>4. write the formula of the anhydrous followed by a dot then the \\# of moles of water and “H2O”\u003Cbr>Example: CuSO4 · 5H2O |  | Naming a Hydrate\u003Cbr>1. write the name of the ionic compound\u003Cbr>2. use a prefix for the number of water molecules (see below)\u003Cbr>3. write “hydrate” after the prefix\u003Cbr>1 = mono, 2 = di, 3 = tri, 4 = tetra, 5 = penta,\u003Cbr>6 = hexa, 7 = hepta, 8 = octa, 9 = nona, 10 = deca\u003Cbr>Example: Copper (II) sulfate pentahydrate |  |  |\n\n\n| Stoichiometry – Interpreting Chemical Equations\u003Cbr>􀁸 “ANY TIME YOU HAVE MOLE AND MOLE IN THE SAME FRACTION ALWAYS LOOK AT THE BALANCED CHEMICAL EQUATION”\u003Cbr>􀁸 Stoichiometry is used to convert from moles of one substance to moles of a different substances |  |\n| --- | --- |\n| o | These substances are related by their mole ratios established by the balanced chemical eq. |\n\n\n| Mole-to-Mole Conversion\u003Cbr>1. start with a balanced chemical equation\u003Cbr>2. start with substance A\u003Cbr>3. multiply by the mole ratio with the substance you want to get “B” on top of the substance you are getting rid of“A”\u003Cbr>4. insert the corresponding coefficient from the equation to the ratio for each substance | \\# mols Substance “A”\u003Cbr>Example: 2H2 + O2 􀃆 2H2O\u003Cbr>\\# mols O2  2 mols H2O 1 mol O2\u003Cbr>\\# mols Substance “A”  \\# mols Substance “B” |\n| --- | --- |\n| Mole-to-Mass Conversion\u003Cbr>1. start information is the same but you\u003Cbr>are asked to find grams of new\u003Cbr>substance\u003Cbr>2. follow steps listed above\u003Cbr>3. multiply by the molar mass of the new\u003Cbr>substance to get the grams of the new\u003Cbr>substance | Example: 2H2 + O2 􀃆 2H2O\u003Cbr>\\# mols O2 2 mols H2O 18 g H2O 1 mol O2 1 mol H2O |\n| Mass-to-Mass Conversion\u003Cbr>1. start information given in grams and\u003Cbr>asked to find grams\u003Cbr>2. 3 step problem, grams substance A to\u003Cbr>m","cbCait9xHVDApPOE","https://ap.wps.com/l/cbCait9xHVDApPOE","pdf",114702,"English","# Calculating Molar Mass\n# Calculating Percent Composition\n# Calculating Empirical Formula\n# Calculating Molecular Formula\n# Calculating the Formula of a Hydrate\n# Naming a Hydrate\n# Stoichiometry: Interpreting Chemical Equations\n# Mole-to-Mole Conversion\n# Mole-to-Mass Conversion\n# Mass-to-Mass Conversion\n# Determining the Limiting Reactant\n# Determining the Amount of Excess Reactant\n# Calculating Percent Yield","[{\"question\":\"How do you calculate molar mass of a compound?\",\"answer\":\"Write the compound’s formula, determine the number of atoms of each element, multiply by each element’s atomic mass, then add all element contributions.\"},{\"question\":\"What is the process for finding an empirical formula?\",\"answer\":\"Convert percent composition to grams (use 100 g if needed), convert grams to moles, divide by the smallest mole value to get the simplest ratio, then multiply to remove fractions and assign the element symbols.\"},{\"question\":\"How do you determine the limiting reactant and percent yield?\",\"answer\":\"Convert reactant amounts to moles and compare moles divided by each reactant’s balanced-equation coefficient; the smaller value indicates the limiting reactant. For percent yield, compute theoretical yield using stoichiometry, divide actual yield by theoretical yield, then multiply by 100.\"}]","Moles & Stoichiometry Cheat Sheet | PDF",1791924920]